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ELEMENTARY  TREATISE 


GEOMETRY, 


SIMPLIFIED    FOR 


BEGINNERS  NOT  VERSED  IN   ALGEBRA, 


PART  I 


CONTAINING 


PLANE   GEOMETRY, 


WITH  ITS  APPLICATION  TO  THE  SOLUTION  OP  PROBLEMS. 


BY  FRANCIS  J.  GRUND. 
L» 


Effete     3E&ftfon,    stereotype* 


BOSTON: 

CHARLES    J.  HENDEE, 

AND 

G.    W.    PALMER     AND     COMPANY, 

1838. 


DISTRICT  OF  MASSACHUSETTS,  TO  WIT: 

District  Clerk's  Office. 

Bjb  it  remembered,  that  on  the  fourth  day  of  December,  A.  D.  1830,  in  the  fifty- 
fifth  year  of  the  Independence  of  the  United  States  of  America,  Francis  J 
Grhnd,  of  the  said  district,  lias  deposited  in  this  office  the  title  of  a  book,  too 
right  whereof  he  claims  as  author,  in  the  words  following,  to  wit : 

Arc  Elementary  Treatise  on  Geometry,  simplified  for  Beginners  not  versed  m 
Algebra.  Part  I,  containing  Plane  Geometry,  with  its  Application  to  the  Solution  of 
Problems.     By  Francis  J.  Grund.    Second  Edition. 

In  conformity  to  the  act  of  the  Congress  of  the  United  States,  entitled,  "  An 
act  for  the  encouragement  of  learning,  by  securing  the  copies  of  maps,  charts, 
and  books,  to  the  authors  and  proprietors  of  such  copies,  during  the  times  therein 
mentioned  ;"  and  also  to  an  act,  entitled,  "  An  act  supplementary  to  an  act, 
entitled, '  An  act  for  the  encouragement  of  learning,  by  securing  the  copies  of 
maps,  charts,  and  books,  to  the  authors  and  proprietors  of  such  copies  during  the 
times  therein  mentioned  ;'  and  extending  the  benefits  thereof  to  the  arts  of  de- 
signing, engraving,  and  etching  historical  and  other  prints." 

').;*-  *     •  •  •  ,'JtooC  W.  DAVIS, 

\     ,.  •derkqf't^e'Di\1rict  of  Massachusetts. 


CAJORI 


J    . 

RECOMMENDATIONS. 


From  John  Farrar,  Professor  of  Mathematics  and  Natural  Philosophy 
at  Harvard   University. 

Mr.  Gruud's  Elementary  Treatise  on  Geometry  contains  much  useful 
matter,  not  generally  to  be  found  in  English  works  of  this  description. 
There  is  considerable  novelty,  also,  in  the  style  and  arrangement.  The 
subject  appears  to  be  developed  in  a  manner  well  suited  to  the  younger 
class  of  learners,  and  to  such  an  extent,  and  with  such  illustrations,  as 
renders  it  a  valuable  introduction  to  the  more  extended  works  on  Ge- 
ometry. 

JOHN  FARRAR. 

February  18th,  1830. 


From  G.  B.  Emerson,  Principal  of  the  English  Classical  School, 
Boston. 
Mr.  Grund's  Geometry  unites,  in  an  unusual  degree,  strictness  of  de- 
monstration with  clearness  and  simplicity.  It  is  thus  very  well  suited  to 
form  habits  of  exact  reasoning  in  young  beginners,  and  to  give  them 
favorable  impressions  of  the  science.  I  have  adopted  it  as  a  text  book 
in  my  own  school. 

GEO.  B.  EMERSON. 
February  18th,  1830. 

»  

From  E.  Bailey,  Principal  of  the  Young  Ladies'  High  School,  Boston. 

Dear  Sir — From  the  specimens  of  your  work  on  Geometry  which  I 
have  seen,  and  especially  from  the  sheets  I  have  used  in  my  school  since 
it  went  to  the  press,  I  have  formed  a  high  opinion  of  its  merits.  The 
general  plan  of  the  work  appears  to  be  very  judicious,  and  you  have 
executed  it  with  great  ability.  Simplicity  has  been  carefully  studied,  yet 
not  at  the  expense  of  rigid  demonstration.  In  this  respect,  it  seems  ad- 
mirably fitted  for  the  use  of  common  schools.    Believing  your  work  cal- 


918317 


4  RECOMMENDATIONS. 

culated  and  destined  to  do  much  good,  in  a  department  of  science  which 
has  been  too  long  neglected,  I  hope  it  may  soon  become  generally  known. 
Very  respectfully,  yours,  &c. 

E.  BAILEY. 
February  17th,  1830. 


From  F.  P.  Leverett,  Principal  of  the  Latin  School,  Boston. 

December  7th,  1830. 
Dear  Sir — I  have  looked  with  much  satisfaction  over  the  sheets  of 
the  second  edition  of  your  '  First  Lessons  in  Plane  Geometry.'  It  is  a 
more  simple  and  intelligible  treatise  on  Geometry  than  any  other  with 
which  I  am  acquainted,  and  seems  to  me  well  adapted  to  the  understand- 
ings of  young  scholars. 

I  am,  dear  sir,  respectfully  yours, 

F.  P.  LEVERETT. 


From  William  B.  Fowle,  Principal  of  the  Monitorial  School,  Boston. 

Boston,  February  17th,  1830. 
Mr.  Grund — Dear  Sir— I  have  examined  every  page  of  your 
'First  Lessons  in  Plane  Geometry.'  Its  reception  everywhere  augurs 
well  for  the  success  of  your  book,  which  is  an  extension  and  practical 
application  of  Fraucceur's.  It  has  fulfilled  my  wishes,  and  I  shall  imme- 
diately introduce  it  into  my  school. 

Yours,  very  respectfully,    - 

•     WILLIAM  B.  FOWLE. 


krom  Walter  R.  Johnson,  Principal  of  the  Philadelphia  High  School. 
Philadelphia,  Nov.  27th,  1830. 
Dear  Sir— The  First  Lessons  in  Plane  Geometry,  with  a  perusal  of 
which  I  have  been  favored,  appears  to  me  eminently  calculated  to  lay 
the  foundation  of  a  clear  and  comprehensive  knowledge  of  the  demon- 
strative parts  of  that  important  science. 

As  it  has  obviously  been  the  result  of  actual  experience  in  teaching,  it 
commends  itself  to  the  attention  of  the  profession,  by  the  assurance  that 
it  is  really  adapted  to  the  comprehension  and  attainments  of  those  for 
whom  it  was  designed.  Pennit  me  to  express  the  hope  that  it  may  meet 
its  full  share  of  that  encouragement  which  works  in  this  department  are 
beginning  to  receive  in  every  part  of  our  country. 

I  remain,  dear  sir,  very  respectfully  yours, 

WALTER  R.  JOHNSON. 


PREFACE. 


Popular  Education,  and  the  increased  study  of  Mathe- 
matics, as  the  proper  foundation  of  all  useful  knowledge, 
seem  to  call  especially  for  Elementary  Treatises  on  Geome- 
try, as  has  been  evinced  in  the  favorable  reception  of  the 
first  edition  of  this  work  within  a  few  months  of  the  date  of 
its  publication.  A  few  changes  have  been  made  in  the 
present  edition,  which,  it  is  hoped,  will  contribute  to  the  use- 
fulness of  the  work  as  a  book  for  elementary  instruction. 

The  author  acknowledges  with  pleasure  the  valuable  aid 
he  has  received  from,  some  of  the  most  experienced  and 
distinguished  instructers ;  and  is,  in  this  respect,  particularly 
indebted  to  the  kindness  of  Messrs.  E.  Bailey,  George  B. 
Emerson,  and  Miss  Elizabeth  P.  Peabody,  of  Boston,  at 
whose  suggestion  several  demonstrations  have  been  simpli- 
fied, in  order  to  adapt  the  work  to  the  capacity  of  early 
beginners. 

As  regards  the  use  of  it  in  schools  and  seminaries,  the 
teacher  will  find  sufficient  directions  in  the  remarks  inserted 
in  the  body  of  the  work. 

The  Problems,  of  which  the  third  and  fourth  parts  are 
principally  selected  from  those  of  Meier  Hirsch,  form  a 
section  by  themselves,  in  order  to  be  more  easily  referred  to. 
1* 


6  PREFACE. 

The  teacher  may,  according  to  his  own  judgment,  use  as 
many  of  them  at  the  end  of  each  section,  as  may  be  solved 
by  the  principles  the  pupils  have  become  acquainted  with. 

Boston,  September  30, 1830. 


PREFACE  TO  THE  STEREOTYPE  EDITION. 

The  present  stereotype  edition  differs  from  the  previous 
ones  only  in  the  typographical  arrangement,  to  meet  the 
view  of  the  publishers,  whose  intention  it  is  to  reduce  its 
price,  in  order  to  bring  it  within  the  reach  of  common  schools 
throughout  the  Union. 

F.J.G. 

Boston,  March  27, 1832. 


^fc' 


TABLE   OF  CONTENTS 


Introduction 9 

Definitions .' 11 

Questions  on  Definitions 15 

.Notation  and  Significations 18 

Axioms {JO 

SECTION  L 

Of  Straight  Lines  and  Angles 22. 

Recapitulation  of  the  Truths  contained  in  the  First  Sec- 
tion   34 

SECTION  II. 

PART  I. 

Of  the  Equality  of  Triangles 37 

PART  II. 

Of  Geometrical  Proportions  and  Similarity  of  Triangles  ....  53 

Theory  of  Geometrical  Proportions 53 

Similarity  of  Triangles 67 

Recapitulation   of  the    Truths    contained   in   Section  IL, 

Part   I 76 

Recapitulation  of  the  Truths  contained  in  Part  II 79 

Questions  on  Proportions 79 

Questions  on  Similarity  of  Triangles 81 


«* 


8  CONTENTS. 

SECTION  III. 

Of  the  Measurement  of  Surfaces 83 

Recapitulation  of  the  Truths  contained  in  the  Third  Sec- 
tion       99 

SECTION  IV. 

Of  the  Properties  of  the  Circle 102 

What  is  meant  by  Squaring  a  Circle 128 

Recapitulation  of  the  Truths  contained  in  the  Fourth  Sec- 
tion   134 

SECTION  V. 

Application   of   the   foregoing   Principles  to  the  So- 
lution of  Problems 141 

PART  I. 

Problems  relative  to  the  Drawing  and  Division  of  Lines 
and  Angles % 141 

PART  II. 
Transformation  of  Geometrical  Figures 159 

PART  III. 
Partition  of  Figures  by  Drawing 171 

PART  IV. 

Construction  of  Triangles t 182 

Appendix,  containing  Exercises  for  the  Slate 189 


GEOMETRY. 


INTRODUCTION. 

If,  without  regarding  the  qualities  of  bodies,  viz  : 
their  smoothness,  roughness,  color,  compactness,  tenacity, 
&c,  we  merely  consider  the  space  which  they  fill — their 
extension  in  space — they  become  the  special  subject  of 
mathematical  investigation,  and  the  science  which  treats 
of  them,  is  called  Geometry.    - 

The  extensions  of  bodies  are,  called  dimensions.  Every 
body  has  three  dimensions,  viz  :  length,  breadth,  and 
depth.  Of  a  wall  or  a  house,  for  instance,  you  can  form 
no  idea,  without  conceiving  it  to  extend  in  length,  breadth, 
and  depth ;  and  the  same  is  the  case  with  every  other 
body  you  can  think  of. 

The  limits  or  confines  of  bodies  are  called  surfaces 
(superfices),  and  may  be  considered  independently  of  the 
bodies  themselves.  So  you  may  look  at  the  front  of  a 
house,  and  inquire  how  long  and  how  high  is  that  house, 
without  regarding  its  depth;  or  you  may  consider  the 
length  and  breadth  of  a  field,  without  asking  how  deep  it 
goes  into  the  ground,  &c.  In  all  such  cases,  you  merely 
consider  two  dimensions.  A  surface  is,  therefore,  defined 
to  be  an  extension  in  length  and  breadth  without  depth. 


16  GEOMETRY. 

The  limits  or  edges  of  surfaces  are  called  lines,  and 
may  again  be  considered  independently  of  the  surfaces 
themselves.  You  may  ask,  for  instance,  how  long  is  the 
front  of  such  a  house,  without  regarding  its  height ;  or 
how  far  is  it  from  Boston  to  Roxbury,  without  inquiring 
how  hroad  is  the  road.  Here,  you  consider  evidently 
only  one  dimension  ;  and  a  line,  therefore,  is  defined  to 
be  an  extension  in  length  without  breadth  or  depth. 

The  beginning  and  end  of  lines  are  called  points 
They  merely  mark  the  positions  of  lines,  and  can,  there- 
fore, of  themselves,  have  no  magnitude.  To  give  an 
example  :  when  you  set  out  from  Boston  to  Roxbury,  you 
may  indicate  the  place  you  .start  from,  which  you  may 
call  the  point  of  starting.  If  this  chances  to  be  Marl- 
borough Hotel,  you  do  not  ask  how  long,  or  broad,  or 
deep  that  place  is ;  it  suffices  for  you  to  know  the  spot 
where  you  begin  your  journey.  A  point  is,  therefore, 
denned  to  be  mere  position,  without  either  length  or 
breadth. 

Remark.  A  point  is  represented  on  paper  or  on  a 
board,  by  a  small  dot.  A  line  is  drawn  on  paper  with  a 
pointed  lead  pencil  or  pen ;  and  on  the  board,  with  a 
thin  mark  made  with  chalk.  The  extensions  of  sur- 
faces are  indicated  by  lines ;  and  bodies  are  represented 
on  paper  or  on  the  board,  according  to  the  rules  of 
perspective. 

Before  we  begin  the  study  of  Geometry,  it  is  neces- 
sary, first,  to  acquaint  ourselves  with  the  meaning  of  some 
terms,  which  are  frequently  made  use  of  in  books  treating 
on  that  science. 


GEOMETRY.  H 

Definitions. 

A  line  is  called  straight,  when  every  part  of  it  lies  in 
the  same  direction,  thus, 


Any  line  in  which  no  part  is  straight,  is  called  a  curve 
line. 


A  geometrical  plane  is  a  surface,  in  which  two  points 
being  taken  at  pleasure,  the  straight  line  joining  them 
lies  entirely  in  that  surface.*  A  surface  in  which  no 
part  is  plane,  is  called  a  curved  surface.  Any  plane  sur- 
face, terminated  by  lines,  is  called  a  geometrical  figure. 

The  simplest  rectilinear  figure,  terminated  by  three 
straight  lines,  is  called  a  triangle. 

A  geometrical  figure,  terminated  by  four  straight  lines, 
is  called  a  quadrilateral — by  5,  a  pentagon — by  6,  a  hexa- 
gon— by  7,  a  heptagon — by  8,  an  octagon,  &c. 

Any  geometrical  figure,  terminated  by  more  than  three 
straight  lines,  is  (by  some  authors)  called  a  polygon^ 

When  two  straight  lines  meet,  they  form  an  angle ;  the 
point  at  which  they  meet  is  called  the  vertex,  and  the 
lines  themselves  are  called  the  legs  of  the  angle.  When 
a  straight  line  meets  another,  so  as  to  make  the  two 
adjacent  angles  equal,  the  angles   are  called  right  an- 

*  The  teacher  can  give  an  illustration  of  this  definition,  by  taking 
anywhere  on  a  piece  of  pasteboard,  two  points  and  joining  them 
by  a  piece  of  stiff  wire.  Then,  by  bending  the  board,  the  wire, 
which  represents  the  line,  will  be  off  the  board,  and  you  have  a 
curved  surface ;  and  by  stretching  the  board,  so  as  to  make  the 
wire  fall  upon  it,  you  have  a  plane. 

t  Legendre  calls  all  geometrical  figures  polygons. 


12 


GEOMETRY. 


gles,  and  the  lines  are  said  to  be  perpendicular  to  each 
other. 


Any  angle  smaller  than  a  right  angle  is  called  acute, 


and  when  greater  than  a  right  angle,  an  obtuse  angle.* 


Two  lines  which,  lying  in  the  same  plane,  and  how- 
ever far  extended  in  both  directions,  never  meet,  are  said 
to  be  parallel  to  each  other. 


When  two  lines,  situated  in  the  same  plane,  are  not  par- 
allel, they  are  either  converging  or  diverging.  Two  lines 
are  said  to  be  converging,  if,  when  extended  in  the  direc- 
tion we  consider,  they  grow  nearer  each  other;  and 
diverging,  if  the  reverse  takes  place. 


Converging. 


Diverging. 


*  Angles  are  measured  by  arcs  of  circles,  described  with  any 
radius  between  their  legs.  Here  the  teacher  may  state,  that  the 
circle  is  divided  into  360  equal  parts,  called  degrees ;  each  degree, 
again,  into  60  equal  parts,  called  minutes;  a  minute,  again,  sub- 
divided into  60  equal  parts,  called  seconds,  &c. ;  and  that  the 
magnitude  of  an  angle  can  thus  be  expressed  in  degrees,  minutes, 
seconds,  &c.  of  an  arc  of  a  circle,  contained  between  its  legs. 


GEOMETRY. 


13 


A  triangle  is  called  equilateral,  when  all  its  sides  are 
equal. 


A  triangle  is  called  isosceles,  when  two  of  its  sides  only 
are  equal. 


A  triangle  is  called  scalene,  when  none  of  its  sides  are 
equal. 


A  triangle  is  also  called  right-angled,  when  it  contains 
a  right  angle ; 


and  oblique-angled,  when  it  contains  no  right  angle. 


A  parallelogram  is  a  quadrilateral  whose  opposite  sides 
are  parallel. 


v_^ 


A  rectangle,  or  oblong,  is  a  right-angled  parallelogram 


14  GEOMETRY. 

A  square  is  a  rectangle  whose  sides  are  all  epual. 


A  rhombus  or  lozenge,  is  a  parallelogram  whose  sides 
are  all  equal. 


A  trapezoid  is  a  quadrilateral  in  which  two  sides  only 
are  parallel. 


A  straight  line  joining  two  vertices,  which  are  not  on 
the  same  side  of  a  geometrical  figure,  is  called  a  diagonal. 

The  side  which  is  opposite  to  the  right  angle,  in  a 
right-angled  triangle,  is  called  the  hypothenuse. 

A  circle  is  a  surface  terminated  on  all  sides  by  a  curve 
line  returning  into  itself,  all  points  of  which  are  at  an 
equal  distance  from  one  and  the  same  point,  called  the 
centre. 


The  curve  line  itself  is  called  the  circumference.     Any 


GEOMETRY.  15 

part  of  it  is  called  an  arc.  A  straight  line,  drawn  from 
the  centre  of  a  circle  to  any  point  of  the  circumference, 
is  called  a  radius.  A  straight  line,  drawn  from  one  point 
of  the  circumference  to  the  other,  passing  through  the 
centre,  is  called  a  diameter.  A  straight  line,  joining  any 
two  points  of  the  circumference,  without  passing  through 
the  centre,  is  called  a  chord. 

The  plane  surface  included  within  an  arc  of  a  circle 
and  the  chord  on  which  it  stands  is  called  a  segment. 

The  arc  of  a  circle  which  stands  on  a  diameter  is 
called  a  semi-circumference.  The  plane  surface  included 
within  a  semi-circumference  and  a  diameter  is  called  a 
semi-circle. 


The  plane  surface  included  within  two  radii  and  an 
arc  of  a  circle  is  called  a  sector.  (See  the  figure,  page 
14.)  If  the  two  radii  are  perpendicular  to  each  other, 
the  sector  is  called  a  quadrant. 

A  straight  line,  which,  drawn  without  the  circle,  and 
however  far  extended  in  both  directions,  meets  the  cir- 
cumference only  in  one  point,  is  called  a  tangent. 

QUESTIONS  ON  DEFINITIONS. 

What  is  that  science  called,  which  treats  of  the  exten- 
sions of  bodies,  considered  separately  from  all  their  other 
qualities  1 

What  are  the  extensions  of  bodies  called  ? 

What  are  the  limits  or  confines  of  bodies  called  T 

How  do  you  define  a  surface  1 


]5  GEOMETRY. 

What  are  the  limits  of  surfaces  called  ? 

How  do  you  define  a  line  ? 

What  are  the  beginning  and  end  of  lines  called  ? 

How  do  you  define  a  point  ? 

How  is  a  geometrical  point  represented? 

How  is  a  line  represented  ?     How  a  surface  ? 

How  do  you  define  a  straight  line  ? 

What  do  you  call  a  line  in  which  no  part  is  straight  ? 

What  is  that  surface  called,  in  which,  when  two  points 
are  taken  at  pleasure,  the  straight  line  joining  them  lies 
entirely  in  it  ? 

What  do  you  call  a  surface  in  which  no  part  is  plane  ? 

What  is  a  plane  surface  called  when  terminated  by  lines? 

By  how  many  straight  lines  is  the  simplest  rectilinear 
figure  terminated  ? 

What  do  you  call  it  ? 

What  do  you  call  a  geometrical  figure  terminated  by 
four  straight  lines  ? 

What,  if  terminated  by  five  straight  lines? 

What,  if  by  six  ?     By  seven  ?     By  eight  ? 

What  are  all  geometrical  figures  terminated  by  more 
than  three  straight  lines  called  ? 

When  two  straight  lines  meet,  what  do  they  form  ? 

What  is  the  point  where  the  lines  meet  called  ? 

What  do  you  call  the  lines  which  form  the  angle  ? 

If  one  straight  line  meets  another,  so  as  to  make  the 
two  adjacent  angles  equal,  what  do  you  call  these  angles  ? 

What  are  the  lines  themselves  said  to  be? 

What  is  an  angle  which  is  smaller  than  a  right  angle 
called? 

What  an  angle  larger  than  a  right  angle  ? 

What  do  you  call  two  lines,  which,  situated  in  the 
same  plane,  and  however  far  extended  both  wavs  never 
meet? 


GEOMETRY.  17 

When  are  two  lines  said  to  be  converging?     When, 
diverging  ? 

When  a  triangle   has  all  its  sides   equal,  what  is  it 
called  ? 

When  two  of  its  sides  only  are  equal,  what  1 

When  none  of  its  sides  are  equal,  what  ? 

What  is  a  triangle  called,  when  it  contains  a  right 
angle  ? 

What,  if  it  does  not  contain  one  ? 

What  is  a  quadrilateral,  whose  opposite  sides  are  par- 
allel, called? 

What  is  a  right-angled  parallelogram  called  ? 

What  is  an  equilateral  rectangle  called  ? 

What,  an  equilateral  parallelogram  ? 

What,  a  quadrilateral  in   which  two  sides  only  are 
parallel  ? 

How  is  a  circle  terminated  ? 

What  is  the  line  called  which  terminates  a  circle  ? 

What  is  any  part  of  the  circumference  called? 

What,  a  straight  line,  drawn  from  the  centre,  to  any 
point  in  the  circumference  T 

What,  a  straight  line  joining  two  points  of  the  circum- 
ference, and  passing  through  the  centre  ? 

What,  a  straight  line  joining  two  points  of  the  circum- 
ference, without  passing  through  the  centre  ? 

What  is  the  plane  surface,  included  within  an  arc  and 
the  chord  which  joins  its  two  extremities,  called  ? 

What  is  that  part  of  the  circumference  called,  which  is 
cut  off  by  the  diameter  ? 

What,  the  plane  surface  within  a  semi-circumference 
and  a  diameter  ? 

What,  the  surface  within  an  arc  of  a  circle  and  the 
two  radii  drawn  to  its  extremities  ? 
2*  • 


18  GEOMETRY. 

What  is  the  sector  called,  if  the  two  radii  are  perpen- 
dicular to  each  other  ? 

What  is  the  name  of  a  straight  line,  drawn  without  the 
circle,  which,  extended  both  ways  ever  so  far,  touches 
the  circumference  only  in  one  point* 


NOTATION  AND  SIGNIFICATIONS. 

For  the  sake  of  shortening  expressions,  and  thereby  to 
facilitate  language,  mathematicians  have  agreed  to  adopt 
the  following  signs : 

=r  stands  for  equal ;  e.  g.,  the  line  AB  —  CD  means, 
that  the  line  AB  is  equal  to  the  line  CD. 

-}-  stands  for  plus  or  more  ;  e.  g.,  the  lines  AB  -}-  CD 
means,  that  the  length  of  the  line  CD  is  to  be  added  to 
the  line  AB. 

—  stands  for  minus  or  less;  e.g.,liue  AB  —  CD  means, 
that  the  length  of  the  line  CD  is  to  be  taken  away  from 
the  line  AB. 

X  is  the  sign  of  multiplication. 

:  is  the  sign  of  division. 

<^  stands  for  less  than ;  e.  g.,  the  line  AB  <^  CD  means, 
that  the  line  AB  is  shorter  than  the  line  CD. 

>  stands  for  greater  than;  e.  g.,  the  line  AB  ]>  CD 
means,  that  the  line  AB  is  longer  than  the  line  CD. 

A  point  is  denoted  by  a  single  letter  of  the  alphabet 
chosen  at  pleasure ;  e.  g.t 

the  point  B. 

A  line  is  represented  by  two  letters  placed  at  the  be- 
ginning and  end  of  it ;  e.  g.t 

A. B 

the  line  AB. 


GEOMETRY 


19 


An  angle  is  commonly  denoted  by  three  letters,  the 
one  that  stands  at  the  vertex  always  placed  in  the  middle ; 

A 

8 


the  angle  ABC  or  CBA.     It  is  sometimes  also  repre- 
sented by  a  single  letter  placed  within  the  angle ;  e.  g., 


the  angle  a. 

A  triangle  is  denoted  by  three  letters  placed  at  the 
three  vertices;  e. g., 


the  triangle  ABC. 

A  polygon  is  denoted  by  as  many  letteis  as  there  are 
vertices ;  e.  g., 


£! 


£> 


the  pentagon  ABCDE. 

A  quadrilateral  is  sometimes  denoted  only  by  two  let- 
ters, placed  at  the  opposite  vertices;  e.  g.t 


the  quadrilateral  AB. 


20  GEOMETRY. 


QUESTIONS  ON  NOTATION  AND  SIGNIFICATIONS. 

What  is  the  sign  of  equality  ? 

What  sign  stands  for  plus  or  more  ? 

What  for  minus  or  less  1 

What  for  multiplication  ? 

What  for  division  1 

What  for  less  than  ? 

What  for  more  than  ? 

How  is  a  point  denoted  ? 

How  a  line  ? 

How  an  angle  ?- 

How  a  triangle  ? 

How  a  quadrilateral  1 

How  any  polygon  1 


Axioms. 

There  are  certain  invariable  truths,  which  are  at  once 
plain  and  evident  to  every  mind,  and  which  are  frequently 
made  use  of,  in  the  course  of  geometrical  reasoning.  As 
you  will  frequently  be  obliged  to  refer  to  them,  it  will  be 
well  to  recollect  the  following  ones  particularly  : 

TRUTH  I. 
Things  which  are  equal  to  the  same  thing,  are  equal 
to  one  another. 

TRUTH  II. 
Things  which  are  similar  to  the  same  thing,  are  similar 
to  one  another. 


-  GEOMETRY.  21 

TRUTH  III. 
If  equals  be  added  to  equals,  the  wholes  are  equal. 

TRUTH  IV. 

If  equals  be  taken  from  equals,  the  remainders  are 

equal. 

TRUTH  V. 

The  whole  is  greater  than  any  one  of  its  parts. 

TRUTH  VI. 
The  sum  of  all  the  parts  is  equal  to  the  whole. 

TRUTH  VII. 
Magnitudes  which  coincide  with  one  another,  that  is, 
which  exactly  fill  the  same  space,  are  equal  to  one  another. 

TRUTH  VIH. 
Between  two  points  only  one  straight  line  can  be  drawn. 

TRUTH  IX. 
The  straight  line  is  the  shortest  way  from  one  point  to 
another. 

TRUTH  X. 
Through  one  point,  without  a  straight  line,  only  one 
line  can  be  drawn  parallel  to  that  same  straight  line. 


SECTION  I. 

OF  STRAIGHT  LINES  AND  ANGLES. 

QUERY  I. 

In  how  many  points  can  two  straight  lines  cut  each  other? 

Answer.  In  one  only. 

Q.  But   could  not  the  two     A 
straight  lines  AB,  CD,  which 
cut  each  other  in  the  point  E,  ^M- 

have  another  point  common;  ^,— — - -~~    '  ^NvM" 

that  is,  could  not  a  part  of  the  Zs  jg 

line  CD  bend  over  and  touch  the  line  AB  in  M  t 

A.  No. 

Q.  Why  not  1 

A.  Because  +here  would  be  two  straight  lines  drawn 
between  the  same  points  E  and  M,  which  is  impossible. 
(Truth  VIII.) 

QUERY  II. 

If  two  lines  have  any  part  common,  what  must  necessa- 
rily follow  ? 

A.  They  must  coincide  with  each  other  throughout, 
and  make  hut  one  and  the  same  straight  line. 

Q.  How  can  you  prove     ,  j£         ^ 

this,  for  instance,  of  the 
two  lines  CA,  BM,  which  have  the  part  MA  common? 

A .  The  common  part  MA  belongs  to  the  line  MB  as 
well  as  to  the  line  AC,  and  therefore  MC  and  AB  are,  in 
this  case,  but  the  continuation  of  the  same  straight  line  AM. 


GEOMETRY. 


23 


M 


E 


QUERY  III. 

How  great  is  the  sum  of  the  two  adjacent  angles,  which 
are  formed  by  one  straight  line  meeting  another ,  taking 
a  right  angle  for  the  measure  ? 

A.  It  is  equal  to  two  right  angles. 

Q.  How  do  you  prove  this 
of  the  two  angles  ADE,  CDE, 
formed  by  the  line  ED,  meet- 
ing the  line  AC,at  the  point  D? 

A.  Because,  if  at  D  you 
erect  the  perpendicular  DM, 
the  two  angles,  ADE  and  CDE,  occupy  exactly  the  same 
space,  as  the  two  right  angles,  ADM  and  CDM,  formed 
by  the  meeting  of  the  perpendicular;  namely,  all  the 
space  on  one  side  of  the  line  AC.     (See  Truth  VII.) 

Q.  Can  you  prove  the  same  of  the  sum  of  the  two 
adjacent  angles,  formed  by  the  meeting  of  any  other  two 
straight  lines  ? 


QUERY  IV 

Wliat  is  the  sum  of  any 
number  of  angles,  a,  b,  c, 
d,  e,  fyc,  formed  at  the 
same  point,  and  on  the  same 
side  of  the  straight  line 
AC,  taking  again  a  right 
angle  for  the  measure  ? 

A.  It  is  also  equal  to  two  right  angles. 

Q.  Why? 

A.  Because,  by  erecting  at  the  point  B  a  perpendicular 
to  AC,  all  these  angles  will  be  found  to  occupy  the  same 
space  as  the  two  right  angles,  made  by  the  perpendicular 
MB. 


24 


GEOMETRY. 


QUERY  V. 

When  two  straight  lines,  AB,  CD,  cut 
each  other,  what  relation  do  the  angles  which 
are  opposite  to  each  other  at  the  vertex  M, 
bear  to  each  other  ? 

A.   They  are  equal  to  each  other. 

Q.  How  can  you  prove  it  1 

A.  Because,  if  you  add  the  same  angle  a, 
first  to  b,  and  then  to  e,  the  sum  will,  in  both 
cases,  be  the  same  ;  namely,  equal  to  two  right  angles ; 
which  could  not  be,  if  the  angle  b  were  not  equal  to  the 
angle  e  (see  Truth  III) ;  and  in  the  same  manner  I  can 
prove  that  the  two  angles,  a  and  d,  are  equal  to  each 
other. 

Q.  If  the  lines  CD,  AB,  are 
perpendicular  to  each  other, 
what  remark  can  you  make  in 
relation  to  the  angles  d,  b, 
ef  a? 

A.  That  each  of  these  an- 
gles is  a  right  angle. 

Q.  And  what  is  the  sum  of 
all  the  angles,  a,  b,  c,  d,  e,  f, 
around  the  same  point,  equal 
to? 

A.    To  four  right  angles. 

Q.  Why? 

A.  Because  if,  through  that 
point,   you  draw  a  perpendicu- 
lar to  any  of  the  lines,   for  in- 
stance the  perpendicular  MN,  to  the  line  OP,  all  the 
angles,  a,  b,  c,  d,  e,f  taken  together,  occupy  the  same 


d 


M 


GEOMETRY. 


25 


space,  which  is  occupied  by  the  four  right  angles,  formed 
by  the  intersection  of  the  two  perpendiculars  MN,  OP. 

QUERY  VI. 

If  a  triangle  has  one  side,  and  the  two  adjacent  angles, 
equal  to  one  side  and  the  tivo  adjacent  angles  of  another 
triangle,  each  to  each,  what  relation  do  these  triangles 
bear  to  each  other  ? 

A.    They  are  equal. 

Q.  Supposing  in.  this  diagram  the  side  a  b  equal  to 
AB ;  the  angle  at  a  equal  to  the  angle  at  A,  and  the  an- 
gle at  b  equal  to  the  angle  at  B  ;  how  can  you.  prove  that 
the  triangle  a  b  c  is  equal  to  the  triangle  ABC  ? 


A.  By  applying  the  side  ah  to  its  equal  AB,  the  side 
ac  will  fall  upon  AC,  and  be  upon  BC  ;  because  the 
angles  at  a  and  A,  b  and  B,  are  respectively  equal ; ,  and 
as  the  sides  ac,  be,  take  the  same  direction  as  the  sides 
AC,  BC,  they  must  also  meet  in  the  same  point  in  which 
the  sides  AC,  BC,  meet ;  that  is,,  the  point  c  will  fall 
upon  C;  and  the  two  triangles  abc,  ABC,  will  coincide 
throughout.  ?  • 

Q.  IVJiat  relation  do  yon  here  discover  between  the 
equal  sides  and  angles  ? 

A.  That  the  equal  angles  -ate  and  C,  are  opposite  to 
the  equal  sides  ab,  AB,  respectively.  ,. 


26 


GEOMETRY. 


QUERY  VII. 

If  two  straight  lines  are  both  perpendicular  to  a  third 
line,  what  relation  must  they  bear  to  each  other  ? 
A.    They  must  be  parallel. 
Q.  Let    us  suppose    the 


;m 


two  lines  AB,  CD,  to  be 
both  perpendicular  to  a 
third  line,  GH ;  how  can 
you  convince  me  that  AB 
and  CD  are  parallel  ? 

A.  Because,  if  you  ex-  ^ 
tend  AB  and  CD,  in  the 
directions  BE,  DF,  making 
BE  and  DF  equal  to  BA 
and  DC  respectively,  every 
thing  will  be  equal  on 
both  sides  of  the  line  GH. 


A 

I        ^ 

i 
O 

B 

D 

M 

F 

II 


jsr 


Now  if  the  l.ines  AB,  CD,  are  not  parallel,  they  must  either 
be  converging  or  diverging.  If  they  are  converging,  AB 
and  CD  wiir,  when  sufficiently  extended,  cut  each  other 
somewhere,  say  in  M  ;  but  then  (every  thing  being  equal 
on  both  sides  of  the  line  GH)  the  same  must  take  place 
with  the  lines  BE,  DF,  on  the  other  side  of  the  line  GH, 
which  must  cut  each  other  somewhere  in  N  ;  and  there 
would  be  two  straight  lines  cutting  each  other  in  two 
points,  which  is  impossible.  If  the  lines  AB,  CD,  were 
diverging,  BE,  DF,  would  be- the  same  ;  but  it  is  equally 
impossible  for  two  straight  lines  to  diverge  in  two  direc- 
tions:  consequently  the  two  straight  lines,  AB,  CD,  can 
neither  be  converging  nor  diverging,  and  therefore  they 
must  be  parallel.. 

Q.   Can  two  straight  lines  which  meet  each  other,  be 
perpendicular  to  the  same  straight  line  ? 


GEOMETRY. 


27 


A.  No. 

Q.  Why  not? 

A.  Because,  if  they  are  both  perpendicular  to  a  third 
line,  I  have  just  proved  that  they  must  be  parallel ;  and 
if  they  are  parallel,  they  cannot  meet  each  other. 

Q.  From  a  point  without  a  straight  line,  how  many 
perpendiculars  can  there  be  drawn  to  that  same  straight 
line  ? 

A.  Only  one.  ' 

Q.  Why  can  there  not  more  be  drawn  ! 

A.  Because  I  have  proved  that  two  perpendiculars  to 
the  same  straight  line  must  be  parallel  to  each  other ;  and 
two  lines,  parallel  to  each  other,  cannot  be  drawn  from 
one  and  the  same  point.  . 


QUERY  VIII. 

If  a  straight  line,  MN, 
cuts  two  other  straight 
lines  at  equal  angles ;  that 
is,  so  as  to  make  the  angles  , 

CIN  and  AFN  equal; 
what   relation    exists    be-  *»' 
tween  these  two  lines  1 

A.  They  are  parallel 
to  each  other. 

Q.  How  can  you  prove 
it  by  this  diagram  ?  The  line  IF  is  bisected  in  O,  and, 
from  that  point  O,  a  perpendicular  OP  is  let  fall  upon  the 
line  AB,  and  afterwards  extended  until,  in  the  point  R, 
it  strikes  the  line  CD. 

A.  I  should  first  observe  that  the  triangles  OPF  and 
ORI  are  equal ;  because  the  triangle  OPF  has  a  side  and 
two  adjacent  angles  equal  to  a  side  and  two  adjacent 
angles  of  the  triangle  ORI,  each  to  each.     (Query  6.) 


Wffft- 


OS  GEOMETRY. 

Q.  Which  is  that  side,  and  which  are  the  two  adjacent 
angles  1 

A.  The  side  OI,  which  is  equal  to  OF  ;  because  the 
point  O  bisects  the  line  IF.  One  of  the  two  adjacent 
angles  is  the  angle  IOR,  which  is  equal  to  the  angle  FOP  \ 
because  these  angles  are  opposite  at  the  vertex  :  and 
the  other  is  the  angle  OIR,  which  is  equal  to  the  angle 
OFP ;  because  the  angle  CIN,  which,  in  the  query,  is 
supposed  to  be  equal  to  AFN,  is  also  equal  to  the  angle 
OIR,  to  which  it  is  opposite  at  the  vertex.     (Truth  I.) 

Q.  But  of  what  use  is  your  proving  that  the  triangle 
ORI  is  equal  to  the  triangle  OPF  ? 

A.  It  shows  that  since  the  triangle  OPF  is  right-an- 
gled in  P,  the  triangle  ORI  must  be  right-angled  in  R ; 
for,  in  equal  triangles,  the  equal  angles  are  opposite  to 
the  equal  sides  (remarks  to  Query  6,  page  25) ;  conse- 
quently the  two  lines  AB,  CD,  are  both  perpendicular  to 
the  same  straight  line  PR,  and  therefore  parallel  to  each 
other.     (Last  query.) 

Q.  Supposing,  now,  two 
straight  lines,  AB,  CD,  to 
be  cut  by  a  third  line,  MN,  so 
as  to  make  the  alternate  an- 
gles AEF  and  EFD,  or  the 
angles  BEF and  EFC, equal, 
what  relation  icould  the  lines 
AB,  CD,  then  bear  to  each  other  ? 

A.    They  icould  still  be  parallel. 

Q.  How  can  you  prove  this?' 

A.  If  the  angle  AEF  is  equal  to  the  angle  EFD,  the 
angles  AEF  and  CFN  are  also  equal ;  because  EFD  and 
CFN  are  opposite  angles  at  the  vertex.  And,  in  the  same 
manner,  it  may  be  proved,  that  if  the  angles  BEF  and 
EFC  are  equal,  MEA  and  EFO  arc  also  equal ;  there- 


GEOMETRY.  29 

fore,  in  both  cases,  there  are  two  straight  lines  cut  by  a 
third  line  at  equal  angles ;  consequently  they  are  parallel 
to  each  other. 

Q.  There  is  one  more  case,  and  that  is :  If  the  two 
straight  lines  AB,  CD  (in  our  last  figure),  are  cut  by  a 
third  line  MN,  so  as  to  make  the  sum  of  the  two  interior 
angles  AEF  and  EF<3y  equal  to  two  right  angles,  how 
are  the  straight  lines  AB,  CD,  then,  situated  with  regard 
to  each  other  ? 

A..  They  are  still  parallel  to  each  other.  For  the  sum 
of  the  two  adjacent  angles  EFC  and  CFN  is  also  equal 
to  two  right  angles  ;  and  therefore,  by  taking  from  each 
of  -the  equal  sums  the  common  angle  EFC',  the  two  re- 
maining angles  AEF  and-  CFN  must  be  equal  (Truth 
IV.) ;  and  you  have  again  the  first  case,  viz  :  two  straight 
lines  cut  by  a  third  line  at  equal  angles. 

Q.  Will  you  now  state  the  different  cases  in  which  two 
straight  lines  are  parallel  1 

A.   1.  When  they  are  cut  by  a  third,  line  at  equal  angles. 

2.  When  they  are  cut  by  a  third  line  so  as  to  make  the 
alternate  angles  equal ;  and, 

3.  When  the  sum  of  the  two  interior  angles,  made  by 
the  intersection  of  a  third  line,  is  equal  to  two  right  ^ 
angles. 


3* 


30 


GEOMETRY. 


QUERY  IX. 

Supposing  the  two  straight  lines  CD,  EFj  are  cut  by 
a  third  line  AM  at  unequal  angles,  ABC,  BHE  (Fig.  I. 
and  II) ;  or  so  as  to  have  the  alternate  angles  CBH  and 
BHF,  or  DBH  and  BHE  unequal ;  or  in  such  a  man- 
ner, that  the  sum  of  the  two  interior  angles  CBH  and 
BHE  (Fig.  I),  or  DBH  and  BHF  (Fig.  II),  is  less 
than  two  right  angles  ;  what  will  then  be  the  case  with  the 
two  straight  lines  CD,  EF?  .   - 

Fig.  I.  .  Fig.  II. 

KCE.  C&       M 


L 


n 


-M 


Dp       F  p*s    F 

A.  They  will,  in  every  one  of  these  cases,  cut  each 
other,  if  sufficiently  extended. 

Q.  How  can  you  prove  this  1 

A.  By  drawing,  through  the  point  B,  another  line  NP 
at  equal  angles  with  EF,  and  which  will  then  also  make 
the  alternate  angles,  NBH,  BHF  and  PBH,  BHE,  equal, 
and  the  sum  of  the  two  interior  angles,  NBH  and  BHE, 
equal  to  two  right  angles ;  this  line  NP  will  be  parallel  to 
the  line  EF ;  consequently  the  line  CD  cannot  be  parallel 
to  it ;  because  through  the  point  B  only  one  line  can  be 
drawn  parallel  to  the  line  EF.     (Truth  X.) 


GEOMETRY. 


31 


QUERY  X. 

Can  you  now  tell  the  relation  which  the  eight  angles } 
a,  b,  c,  d,  e,f,  g,  h,  formed  by  the  intersection  of  two  par- 
allel lines,  by  a  third  line,  bear 
to  each  other  ? 

A.  Yes.  In  the  first  place, 
the  angle  a  is  equal  to  the  angle 
e ;  the  angle  c  equal  to  the  angle 
g ;  the  angle  b  equal  to  the  angle 
f ;  and.  the  angle  d  equal  to  the 
angle  h; — 2d.  the  angles  a,  d-, 
e,  h,  as  well  as  the  angles  b,  c,  f,  g,  are  respectively  equal 
to  one  another  ; — and  finally,  the  sum  of  either  c  and  e,  or 
d  and  f,  must  make  two  right  angles.  For  if  either  of 
these  cases  were  not  true,  the  lines  would  not  be  parallel, 
(Last  query.) 


QUERY  XE     . 

From  what  you  have,  learned  of  the  properties  of  paral- 
lel lines,  what  law  do  you  discover  respecting  the  distance 
they  keep  from  each  other? 

A.  Parallel  lines  remain  throughout  equidistant. 

Q.  When  do  you  call  two  lines  equidistant? 

A .  When  all  the  perpendiculars,  let  -fall  from  one  line 
upon  the  other,  are  equal. 

Q.  How  can  you  prove,  that  the  perpendicular  lines 
OP,  MI,  RS,  &c.  are  all  equal  to  one  another  1 

M  O  £ 


A- 


& 


I 


D 


A.  By  joining  MP,  the  two  triangles  MPO,  MPI,  have 
the  side  MP  common ;  and  the  angle  a  is  equal  to.  the 


32  GEOMETRY. 

angle  b ;  because  a  and  b  are  alternate  angles,  formed  by 
the  two  parallel  lines  MI,  OP  (Query  10) ;  and  the  angle 
c  is  equal  to  the  angle  d;  because  these  angles  are  formed 
in  a  similar  manner  by  the  parallel  Hues  AB,  CD  :  there- 
fore we  have  a  side  and  two  adjacent  angles  in  the  trian- 
gle MPO,  equal  to  a  side  and  two  adjacent  angles  in  the 
triangle  MPI ;  consequently  these  two  triangles  are  equal ; 
and  the  side  OP,  opposite  to-  the  angle  c,  in  the  triangle 
MPO,  is  equal  to  the  side  MI,  opposite  to  the  equal  angle 
d,  in  the  triangle  MPI.  In  precisely  the  same  manner  I 
can  prove  that  RS  is  equal  to,  MI,  and  consequently  also 
to  OP;  and  so  I  might  go  on,  and  show  that  every  per- 
pendicular, let  fall  from  the  line  AB,  upon  the  parallel 
line  CD,  is  equal  to  RS,  MI,  OP,  &c-  The  two  parallel 
lines*  AB  and  CD  are  therefore,  throughout,- at  ah  equal 
distance  from  each  other ;  and  the  same  can  be  proved 
of  other  parallel  lines. 

QUERY  XII. 

If  two  lines  are  parallel  to  a  third  line,  what  relation 
do  they  bear  to  each  other  ?  . 

Fig.  I. 
C- i , D 

a — '■ — —I 1 — : — —b 


F 


Fig,  II. 

A. 1 j -B 

C 1 1 D 

E ! ■ F 

They  are  parallel  to  each  other. 

Q.  How  can  you  prove  this  1 

A.  From  the  line  CD  being  parallel  to  AB,  it  follows 
that  every  point  in  the  line  CD  is  at  an  equal  distance 
from  the  line  AB ;  and  because  EF  is  also  parallel  to  AB, 


GEOMETRY. 


every  point  in  the  line  EF  is  also  at  an  equal  distance 
from  the  line  AB  ;  and  therefore  (in  Fig.  I.)  the  whole 
distances  between  the  lines  CD  and  EF,  or  (in  Fig.  II.) 
the  differences  between  the  equal  distances,  are  equal : 
that  is,  the  lines  CD,  EF,  are  likewise  equidistant ;  and 
consequently  parallel  to  each  other. 

QUERY  XIII. 

What  is  the  sum  of  all  the  angles  in  every  triangle 
equal  to  ? 

A.   To  two  right  angles, 

Q.  How  do  you  prove  this  1 

A. By  drawing,  through  the 
vertex  of  the  angle  b,  a  straight 


line  parallel  to  the  basis  BC,  the 


^(f 


angle  a  is  equal  to  the  angle  d,  and  the  angle  c  is  equal 
to  the  angle  c  (Query  10) ;  and  as  the  sum  of  the  three 
angles  ay  br  c,  is  equal  to  two  right  angles  (Query  4),  the 
sum  of  the  three  angles  d,  bf  e,  in  the  triangle,  is  also 
equal  to  two  right  angles.** 

Q.  Can  you  now  find 
out  the  relation  which  the 
exterior  angle  e  bears  to 
the  two  interior  angles  a 
and  hi  •     ' 

A.  The  exterior  angle 
e  is  equal  to  the  sum  of 
the  two  interior  angles ,  a  and  b. 

Q.  How  can  you  prove  this  ? 

A.  Because,  by  adding  the  angle  c  to  the  two  angles  a 

*  The  teacher  may  give  his  pupils  an  ocular  demonstration  of 
this  truth,  by  cutting  the  three  angles  b,  d,  e,  from  a  triangle,  and 
then  placing  them  along  side  of  each  other ;  they  will  be  in  a 
straight  line. 


34  GEOMETRY. 

and  b,  it  makes  with  them  two  right  angles ;  and  by  add- 
ing it  to  the  angle  e  alone,  the  sum  of  the  two  angles, 
c  and  c,  is  also  equal  to  two  right  angles  (Query  3),  which 
could  not  be,  if  the  angle  e  alone  were  not  equal  to  the 
two  angles  a  and  b  together.     (Truth  III.) 

Q.  What  other  truths  can  you  derive  from  the  two 
which  you  have  just  now  advanced? 

A.  1.  The  exterior  angle  e  is  greater  than  either  of  the 
interior  opposite  ones,  a  or  b. 

2.  If  two  angles  of  a  triangle  are  known,  the  third 
angle  is  also  determined. 

3.  Wlien  .two  angles  of  a  triangle  are  equal  to  two 
angles  of  another  triangle,  the  third  angle  in  the  one  is 
equal  to  the  third  angle  in  the  other. 

4.  No  triangle  can  contain  more  than  one  right  angle. 

5.  No  triangle  can  contain  more  than  one  obtuse  angle. 

6.  No  triangle  can  contain  a  right  and -an  obtuse  angle 
together. 

7.  In  a  right-angled  triangle,  the  right  angle  is  equal 
to  the  sum  of  the  two  other  angles* 

Q.  How  can  you  convince,  me  of  the  truth  of  each  of 
these  assertions  1 


RECAPITULATION     OF    THE    TRUTHS    CONTAINED    IN 
THE    FIRST    SECTION. 

Can  you  now  repeat  the  different  principles  of  straight 
lines  and  angles  which  you  have  learned  in  this  section? 

Ans.  1.  Two  straight  lines  can  cut  each  other  only  in 
one  point. 

•  2..  Two  straight  lines  which  have  two  points  common, 
must  coincide  with  each  other  throughout,  and  form  but 
one  and  the  same  straight  line. 


GEOMETRY.  35 

3.  The  sum  of  the  two  adjacent  angles,  which  one 
straight  line  makes  with  another,  is  equal  to  two  right 
angles. 

4.  The  sum  of  all  the  angles,  made  by  any  number  of 
straight  lines,  meeting  in  the  same  point,  and  on  the  same 
side  of  a  straight  line,  is  equal  to  two  right  angles. 

5.  Opposite  angles  at  the  vertex  are  equal. 

6.  The  sum  of  all  the  angles,  made  by  the  meeting  of 
ever  so  many  straight  lines  around  the  same  point,  is 
equal  to  four  right  angles.        -  .  ■  S    ■    • 

7.  When  a  triangle  has  one  side  and  the  two  adjacent 
angles,  equal  to  one  side  and  the  two  adjacent  angles 
in  another  triangle,  each  to  each,  the  two  triangles  are 
equal. 

8.  In  equal  triangles  the  equal  angles  are  opposite  to 
the  equal  sides.  '  .  • 

9.  If  two  straight  lines  are  perpendicular  to  a  third 
line,  they  are  parallel  to  each  other. 

10.  If  two  lines  are  cut  by  a  third  line  at  equal  angles, 
or  so  as  to  make  the  alternate  angles  equal,  or  so  as  10 
make  the  sum  of  trie  two  interior  angles  formed  by  the 
intersection  of  a  third  line,  equal  to  two  right  angles,  the 
two  lines  are  parallel. 

11.  If  two  lines  are  cut;  bv  a  third  line  at  unequal 
angles ;  or  so  as  to  have  the  alternate  angles  unequal ;  or 
in  such  a  way  as  to  make  the  sum  of  the  two  'interior 
angles  less  than  two  right  angles,  these  two  lines  will, 
when  sufficiently  extended,  cut  each  other. 

12.  If  two  parallel  lines  are  cut  by  a  third  line,  the 
alternate  angles  are  equal. 

13.  Parallel  lines,  are  throughout  equidistant. 

14.  If  two  lines  are  parallel  to  a  third  line,  they  are 
parallel  to  each  other. 


36  GEOMETRY. 

15.  The  sum  of  the  three  angles  in  any  triangle,  is 
equal  to  two  right  angles. 

16.  If  one  of  the  sides  of  a  triangle  is  extended,  the 
exterior  angle  is  equal  to  the  sum  of  the  two  interior 
opposite  angles. 

17.  The  exterior  angle  is  greater  than  either  of  the 
interior  opposite  ones. 

18.  If  two  angles  of  a  triangle  are  given,  the  third  is 
determined.  -. 

•19.  There  can  be  but  one  right  angle,  or  one  obtuse 
angle,  and  never  a  right  angle  and  obtuse  angle  together, 
in  the  same  triangle. 

20.  In  a  right-angled  triangle,  the  right  angle  is  equal 
to  the  sum  of  the  two  other  angles.* 


*  The  teacher  may  now  ask  his  pupils  to  repeat  the  demonstra- 
tions of  these  principles. 


SECTION  II. 

OF  EQUALITY  AND  SIMILARITY  OF  TRIANGLES. 


PART  I. 

OF  THE  EQUALITY  OF  TRIANGLES. 

r 

Preliminary  Remark.  There  are  three  kinds  of  equality  to  be 
considered  in  triangles,  viz  :  equality  of  area,  without  reference  to 
the  shape;  equality  of  shape,  without  reference  to  the  area — simi- 
larity ;  and  equality  of  both  shape  and  area — coincidence.  All 
questions,  asked  in  this  section,  will  refer  only  to  the  last  two  kinds 
of  equality ;  and  those  in  the  first  part,  only  to  the  coincidence  of 
triangles. 

UUERY  I. 

If  two  sides  and  the  angle  which  is  included  by  them  in 
one  triangle,  arc  equal  to  two  sides  and  the  angle  which  is 
included  by  them  in  another  triangle,  each  to  each,  what 
relation  do  these  two  triangles  bear  to  each  other  ? 

Ans.  They  are  equal  to  each  other  in  all  their  parts, 
that  is,  they  coincide  with  each  other  throughout. 

Show  me  that  this  must  be  the  case  with  any  two  tri- 
angles, ABC,  abc,  in  which  we  will  suppose  the  side 
AB  —  ab,  AC  =  ac,  and  the  angle  at  A  equal  to  the  angle 
at  a. 

4 


38  GEOMETRY. 


A.  By  placing  the  line  ac  upon  its  equal  AC,  the  angle 
at  a  will  coincide  with  the  angle  at  A,  because  these  two 
angies  are  equal ;  and  the  line  ah  will  fall  upon  the  line 
AB ;  and  as  ah  =  AB,  the  point  h  will  fall  upon  B ;  that 
is,  the  three  points  of  the  triangle  ahc  will  fall  upon  the 
three  points  of  the  triangle  ABC,  thus : 
The  point  a  upon  A, 
"     h      "     B, 
"     c      "     C; 
consequently  these  two  triangles  must  coincide. 

Q.  What  remark  can  you  here  make  with  respect  to  the 
sides  and  angles  of  equal  triangles  ? 

A.  The  equal  sides,  cb,  CB,  are  opposite  to  the  equal 
angles  at  a  and  A. 

QUERY  II. 

If  one  side  and  the  two  adjacent  angles  in  one  triangle, 
are  equal  to  one  side  and  the  two  adjacent  angles  in 
another  triangle,  each  to  each,  what  relation  do  the  two 
triangles  hear  to  each  other  ? 

A.  They  are  equal,  and  the  angles  opposite  to  the  equal 
sides  are  also  equals  as  has  been  proved  in  the  1st  Section. 
(Query  6.) 

QUERY  III. 

What  remark  can  you  make  with  respect  to  the  two 
angles  at  the  basis  of  an  isosceles  triangle  ? 
A.   They  are  equal  to  each  other. 


GEOMETRY.  39 

Q.  How  can  you  prove  it  ? 
A.  Suppose  we  had  two  equal 
isosceles  triangles,  ABC  and 
abc,  or,  as  it  were,  another  im- 
pression, abc,  of  the  triangle 
ABC,  that  is, 

The  side  ab  =  AB, 
«    ac  =z  AC, 
"    be  =zBC. 
The  angle  at  a  sd  angle  at  A, 
b=     "  B, 

cz=     "  C. 

Then  the  sides  AB,  AC,  ab,  ac,  being  all  equal  to  one 
another,  and  the  angle  at  a  remaining  the  same,  which- 
ever way  we  place  it,  the  whole  of  the  two  triangles,  abc, 
and  ABC,  will  still  coincide,  when  abc  is  placed  upon 
ABC  in  such  a  manner  that  ac  will  fall  upon  AB,  and  ab 
upon  AC  (for  you  will  still  have  two  sides  and  the  angle 
which  is  included  by  them  in  the  one,  equal  to  two  sides 
and  the  angle  which  is  included  by  them  in  the  other) ; 
therefore  the  angle  at  c  must  be  equal  to  the  angle  at  B. 
And  as  the  angle  at  c  is  only,  as  it  were,  another  impression 
of  the  angle  at  C,  the  angles  C  and  B  must  also  be  equal ; 
that  is,  the  two  angles  at  the  basis  of  the  isosceles  triangle 
ABC  are  equal :  and  the  same  can  be  proved  of  the  two 
angles  at  the  basis  of  every  other  isosceles  triangle. 

QUERY  IV. 

If  the  three  sides  of  one  triangle  are  equal  to  the  three 
sides  of  another,  each  to  each,  what  relation  do  the  two 
triangles  bear  to  each  other  ? 

A.  Tltey  coincide  with  each  other  throughout ;  that  is, 
their  angles  are  also  equal,  each  to  each. 


40 


GEOMETRY. 


Q.  How  can  you  prove  this,  for 
instance,  of  the  two  triangles  ABC 
and  abc,  in  which  we  will  suppose 
the  side         AB  =  ab, 

AC  =  ac, 

BC  —  bc? 
That  you  may  easier  find  out  your 
demonstration,  I  have  placed  the 
two  triangles,  as  you  see,  along 
side  of  each  other,  with  their  bases,  AB  and  ab}  together, 
and  have  joined  their  opposite  vertices,  C  and  c  by  the 
straight  line  Cc.  What  do  you  now  observe  with  regard 
to  the  two  triangles  ACc  and  BCc  ? 

A.  Both  are  isosceles ;  for  the  sides  AC  and  ac,  BC 
and  be,  are  respectively  equal ;  and,  therefore,  the  angles 
*  and  y,  o  and  w,  must  be  equal,  each  to  each ;  and  as 
the  angle  x  is  equal  to  the  angle  y,  and  the  angle  o  equal 
to  the  angle  w,  the  sum  of  the  two  angles  x  and  o,  that 
is,  the  whole  angle  ACB,  must  be  equal  to  the  sum  of  the 
two  angles  y  and  w,  that  is,  to  the  whole  angle  acb  ;  and 
the  two  triangles,  ABC,  and  abc,  having  two  sides,  AC, 
BC,  and  the  angle  which  is  included  by  them  in  the  one, 
equal  to  the  two  sides  ac,  be,  and  the  angle  which  is  in- 
cluded by  them  in  the  other,  each  to  each,  must  coincide 
throughout,  and  have,  consequently,  all  their  angles  re- 
spectively equal  to  one  another.     (Query  1,  Sect.  II.) 


QUERY  V. 

WJiich  of  two  angles  in  a  triangle  is  greater,  that 
which  is  opposite  to  the  smaller,  or  that  which  is  opposite 
to  the  greater  side  ? 

A.    That  which  is  opposite  to  the  greater  side. 

Q.  How  can  you  prove  it  ? 

A.  Because    if   in  any  triangle,  for  instance  in  the 


GEOMETRY.  41 

triangle  ABD,  one  side,  AB,  is  j> 

greater  than  another,  AD,  the  side 
AB  will  contain  a  part  which  is 

equal  to  AD  ;#  and  therefore,  by  A-     {j^-^ 

taking  upon  AB  the  distance  AC 

equal  to  AD,  and  joining  DC,  the, triangle  ACD  will  be 
isosceles,  and  the  angle  x  will  be  equal  to  the  angle  y, 
(Query  3,  Sect.  II.)  ;  and  as  the  exterior  angle  y  must 
be  greater  than  the  interior  opposite  angle  CBD,  in  the 
triangle  DBC,  (Query  13,  Sect.  J.)  the  angle  at  x  will 
also  be  greater  than  the  angle  CBD ;  and  the  angle  ADB 
being  greater  still  than  the  angle  x,  must  consequently 
be  still  more  so  than  the  angle  CBD ;  that  is,  the  angle 
ADB,  opposite  to  the  greater  side  AB,  is  greater  than  the 
angle  at  B,  opposite  to  the  smaller  side  AD  :  and  the  same 
can  be  proved  of  two  unequal  sides  in  any  other  triangle. 

Q.  What  truth  can  you  directly  derive  from  this,  re- 
specting the  three  angles  and  sides  of  a  triangle  ! 

A.  That  the  greatest  of  the  three  angles  is  opposite  to 
the  greatest  of  the  three  sides.  For  if  the  side  AD,  for 
instance,  is  greater  than  the  side  DB,  it  can  be  proved 
that  the  angle  at  B,  opposite  to  the  side  AD,  is  greater 
than  the  angle  at  A,  opposite  to  the  side  DB  ;  and  as  the 
side  AB  is  greater  still  than  AD,  the  angle  ADB,  opposite 
to  AB,  must  be  greater  still  than  the  angle  at  B,  and  is 
therefore  the  greatest  angle  in  the  triangle  ABD. 

Q.  From  ichat  you  have  learned  of  the  relation  which 
exists  between  the  sides  and.  angles  of  a  triangle,  can  you 
now  tell  which  of  the  sides  of  a  right-angled  triangle  is 
the  greatest  ? 

A.   Yes.     That  which  is  apposite  to  the  right  angle. 

Q.  Why? 

*  If  the  magnitude  A  is  greater  than  B,  A  must  contain  a  part 
equal  to  B. 

4* 


GEOMETRY. 


A.  Because,  in  a  right-angled  triangle,  the  right  angle 
is  greater  than  either  of  the  two  other  angles.  (Conseq. 
Query  13,  Sect.  I.) 

QUERY  VI. 

It  has   been   proved  before   (Query  3, 
Sect.  II.),  that  in  an  isosceles  triangle,  the  A 

angles  at  the  basis  are  equal :  can  you  now  j  \ 

prove  the  reverse;  that  is,  that  a  triangle         /       \ 
must  be  isosceles  ivhen-  it  contains  two  equal       /  \ 

angles?  A  B 

A.  Yes.  Because,  if  either  of  the  two 
sides  AC,  BC,  were  greater  than  the  other,  the  angle 
opposite  to  that  side  would  also  be  greater  than  the  angle 
which  is  opposite  to  the  other  side  ;  but  the  two  angles 
at  A  and  B  are  equal,  therefore  the  sides  AC,  BC,  are 
also  equal. 

Q.  If  the  three  angles  in  a  triangle  are  equal  to  one 
another,  what  relation  do  the  sides  bear  to  each  other  ? 

A.   They  are  also  equal,  and  the  triangle  is  equilateral. 

Q.  How  can  you  prove  this  ? 

A.  If,  in  the  triangle  ABC,  for 
instance,  the  angle  at  A  is  equal 
to  the  angle  at  B,  I  have  just 
proved  that  the  side  BC  must  be 
equal  to  the  side  AC  ;  and  if  the 
angle  at  B  is  also  equal  to  the 
angle  at  C,  the  side  AC  must  likewise  be  equal  to  the 
side  AB ;  that  is,  the  three  sides  AB,  BC,  AC,  are  equal 
to  one  another,  and  the  triangle  ABC  is  equilateral. 

QUERY  VH. 

Can  any  one  side  of  a  triangle  be  greater  than,  or  equal 
to,  the  sum  of  the  two  other  sides  ? 


GEOMETRY.  43 

A.  No.  A  straight  line  being 
the  shortest  way  from  one  point 
to  another,  it  follows  that,  in  any 
triangle,  ABC  for  instance,  the 
side  AB  is  smaller  than  the  sides  AC  and  BC  together. 

QUERY  VIII. 

If,  from  a  point  M,  in  a  triangle  ABC,  two  lines,  AM, 
BM,  are  drawn  to  the  two  extremities  of  any  side,  AB, 
in  that  triangle,  what  relation  does  the  angle  AMB,  made 
by  these  two  lines,  bear  to  the  angle  ACB,  which  is  opposite 
to  the  side  AB  in  the  triangle  ?  And  what  do  you  ob- 
serve with  regard  to  the  sum  of  the  two  lines,  AM  and 
MB,  which  include  the  angle  AMB,  and  that  of  the  two 
sides,  AC,  BC,  of  the  triangle  which  include  the  angle 
ACB? 

C 


DM 


D 
DM 


A  MB 

A.  The  angle  AMB,  made  by  the  lines  AM,  BM,  is 
always  greater  than  the  angle  ACB,  opposite  to  the  side 
AB,  in  the  triangle  ABC;  but  the  sum  of  the  two 
lines  AM,  MB,  is  in  all  cases  smaller  than  the  sum  of  the 
two  sides  AC,  CB,  of  the  triangle. 

Q.  How  can  you  prove  both  your  assertions  I 
A.  The  exterior  angle  MDB  is  greater  than  the  inte- 
rior opposite  angle  ACD,  in  the  triangle  ACD  (Query 
13,  Sect.  I.) ;  and  for  the  same  reason  is  the  exterior 
angle   AMB  greater   than   the   interior   opposite   angle 


44  GEOMETRY. 

MDB,  in  the  triangle  MDB;  and  therefore  the  angle 
AMB  is  greater  still  than  the  angle  ACB.  Idly.  The 
three  sides  AB,  AC,  BC,  by  which  the  greater  surface 
is  bound,  enveloping  the  three  sides  AB,  AM,  MB,  it 
follows  that  their  sum  is  greater  than  the  sum  of  the 
three  sides  AB,  AM,  BM,  by  which  the  smaller  surface 
ABM  is  bound ;  and,  taking  from  each  of  the  unequal 
sums  the  same  line  AB,  which  serves  both  as  a  common 
basis,  the  greater  will  remain  where  the  greater  was 
before ;  that  is,  the  sum  of  AC  and  BC  will  still  be 
greater  than  the  sum  of  AM,  BM. 

QUERY  IX. 

If,  from  a  point  A ,  without  a  straight  line  MN,  you 
let  fall  a  perpendicular,  AB,  upon  that  line;  and,  at  the 
same  time,  draw  other  lines,  AD,  AE,  AF,  fyc,  obliquely 
to  different  points,  I),  E,  F,  fyc,  in  the  same  straight 
line;  which  is  the  shortest,  the  perpendicular,  or  one  of 
the  oblique  lines  I 

A. 


A.   The  perpendicular  is  the  shortest. 

Q.  How  can  you  prove  it  1 

A.  Because  the  triangles,  ABD,  ABE,  ABF,  ABN, 
&c.  are  all  right-angled  in  B ;  and  in  every  right-angled 
triangle,  the  greatest  side  is  opposite  to  the  right  angle. 
(Page  41.) 

Q.  And  what  other  truths  do  you  derive  from  the  one 
you  have  just  mentioned? 

A.  1st.   The  perpendicular  AB  measures  the  distance 


GEOMETRF.  45 

of  the  point  A  from  the  line  MN;  for  it  is  the  shortest 
line  that  can  be  drawn  from  that  point  to  that  line* 

2dly»  The  angles  o,  p,  r,  t,  Sfc.  are  all  obtuse,  because 
they  are  exterior  angles  of  the  right-angled  triangles, 
ABD,  ABE,  ABF,  &C;,  and,  therefore,  greater  than  the 
interior  opposite  right  angle  at  B. 

3dly.  The  angles  o,  p,  r,  t,  8$c.  become  successively 
greater,  and  the  angles  u,  q,  s,  &o.  smaller,  as  the  lines 
AD,  AE,  AF,  fyc,  are  drawn  farther  from  the  perpen- 
dicular. For  the  exterior  angle  p  is  greater  than  the 
interior  opposite  one  o,  in  the  triangle  ADE ;  the  exterior 
angle  r  is  greater  than  the  interior  opposite  one  p,  in  the 
triangle  AEF  ;  the  exterior  angle  t,  again,  is  greater  than 
the  interior  opposite  one  r,  in  the  triangle  AFN ;  and 
so  on. 

4thly.  The  oblique  lines,  AD,  AE,  AF,  fyc.  become 
successively  greater,  as  they  are  drawn  farther  from  the 
perpendicular;  that  is,  the  line  AD  is  greater  than  the 
line  AB ;  the  line  AE  than  the  line  AD ;  the  line  AF 
than  the  line  AE  ;  and  so  on.  For  the  angles  o,  p,  r,  &,c. 
are  all  obtuse,  and  become  successively  greater,  as  the 
triangles  ADE,  AEF,  &c.  are  more  remote  from  the  per- 
pendicular ;  and,  therefore,  the  sides  AE,  AF,  AN,  &c, 
which  are  successively  opposite  to  these  angles,  in  the 
triangles  ADE,  AEF,  AFN,  must  become  greater  with 
them. 

5thly.  The  straight  lines,  AC,  AD,  drawn  on  both 
sides  of,  and  at  an  equal  distance  from,  the  perpendicular 
AB,  are  equal.  For  the  two  triangles  ABC,  ABD,  have 
the  side  AB  common,  and  the  side  BC  equal  to  the  side 
BD  (because  the  lines  AC,  AD,  are  at  an  equal  distance 
from  the  perpendicular  AB) ;  and  as  the  line  AB  is  per- 
pendicular to  CD,  the  angle  ABC,  included  by  the  sides 
AB,  BC,  in  the  triangle  ABC,  is  equal  to  the  angle  ABD, 


46  -  GEOMETRY. 

included  by  the  sides  AB,  BD,  in  the  triangle  ABD ;  con- 
sequently, these  two  triangles  are  equal ;  and  the  third 
side  AC  in  the  one  triangle,  is  equal  to  the  third  side 
AD  in  the  other.     (Query  1,  Sect.  II.) 

6thly.  There  is  but  one  point  in  the  line  MN,  on  each 
side  of  the  perpendicular,  such,  that  a  straight  line, 
drawn  from  it  to  the  point  A,  is  of  a  given  length.  This 
follows  from  No.  4. 

7thly.  There  is  but  one  point  in  the  line  MN,  on  each 
side  of  the  perpendicidar ,  in  which  a  line  drawn  to  the 
point  A  forms  with  the  line  MN  an  angle  of  a  given 
magnitude.     This  follows  from  No.  3. 

QUERY  X. 

If  two  sides,  and  the  angle  which  is  opposite  to  the 
greater  of  them,  in  one  triangle,  are  equal  to  two  sides  ana 
the  angle  which  is  opposite  to  the  greater  of  them  in 
another,  each  to  each,  what  relation  do  these  two  trian* 
gles  bear  to  each  other  ? 

A.  They  coincide  w'ith  each  other  in  all  their  parts; 
that  is,  they  are  equal  to  each  other. 

Q.  How  can  you  prove  it? 
I      A.  Because,   if,    in  a  triangle,  Fig'  L 

ABC,  for  instance,  you  have  the  -£*• 

sides  AB  and  AC,  and  the  angle  / j  \Snss. 

at    B,    which   is   opposite   to   the  nZ. — ! — !^ _^w 

greater  side  AC,  given,  the  whole 

triangle  is  determined.     For,  in  the  jy 

first  place,  by  the  angle  at  B,  the        / 1\^"^^ 

direction  of  the  sides  AB,  BC,  is    jg }"~jj >  jg  :^tf 

determined.     %dly.  By  the  length 

of  the  side  AB,  the  distance  of  the  point  A  from  the  line 

BC  is  determined.     2dly.  If  you  imagine  the  perpendic- 


GEOMETRY.  47 

ular  AD  to  be  let  fall  upon  BC  (Fig.  I.),  or  if  the  angle 
ABC  be  obtuse  (as  in  Fig.  II.)  on  its  further  extension 
BE,  there  can  be  but  one  point  in  the  line  BC,  on  this 
side  of  the  perpendicular,  from  which  a  line  drawn  to  the 
point  A,  is  as  long  as  the  line  AC  (see  consequence  6th 
of  the  preceding  query)  ;  therefore,  by  the  length  of  the 
line  AC,  the  point  C,  and  thereby  the  whole  of  the  third 
line  BC,  is  also  determined. 

Q.  But  is  it  not  possible  for  the  line  AC  to  fall  on  the 
other  side  of  the  perpendicular  ? 

A.  No.  Because  the  line  AC,  being  greater  than  the 
line  AB,  would  in  this  case  be  farther  from  the  perpen- 
dicular, than  the  line  AB  (conseq.  4,  preceding  query), 
and  the  angle  at  B  would  then  fall  without  the  triangle  ; 
and  because  the  whole  triangle  ABC  is  entirely  deter- 
mined, when  two  of  its  sides,  and  the  angle  which,  is 
opposite  to  the  greater  of  them,  are  given  :  therefore,  all 
triangles,  in  which  these  three  things  are  equal,  must  be 
equal  to  one  another. 

Q.  What  truth  can  you  infer  from  this  respecting  the 
case  where  the  hypothcnuse,  and  one  side  of  a  right-angled 
triangle,  are  equal  to  the  hypothcnuse  and  one  of  the 
sides  in  another  right-angled  triangle  ? 


A.  That  these  two  right-angled  triangles  are  equal  to 
each  other.  For,  in  this  case,  we  have  two  sides,  and 
the  right  angle  which  is  opposite  to  the  greater  of  them, 
in  the  one,  equal  to  two  sides,  and  the  angle  which  is 
opposite  to  the  greater  of  them,  in  the  other. 


48  GEOMETRY. 

Q.  But  if,  in  Fig.  II.  (page  46)  the  two  sides  AC,  AB,  and  the 
angle  at  C,  opposite  to  the  smaller  side  AB,  be  given,  would  not 
this  be  sufficient  to  determine  the  triangle  ABC  ? 

A.  No.  For  the  two  lines,  AB,  AE,  being  equal,  there  would 
be  two  triangles,  ABC  and  AEC  possible,  containing  the  same 
three  things,  and  it  would  be  doubtful  which  of  the  two  triangles 
was  meant. 

QUERY  XI. 

If  you  have  two  sides,  ab,  be,  of  a  triangle,  abc,  equal 
to  two  sides,  AB,  BC,of  another  triangle,  ABC,  each  to 
each  ;  hut  the  angle  ABC  included  by  the  two  sides,  AB, 
BC,  in  the  triangle  ABC,  greater  than  the  angle  abc, 
included  by  the  sides  ab,  be,  in  the  triangle  abc ;  ichat 
remark  can  you  make  with  regard  to  the  two  sides  ac, 
AC,  which  are  respectively  opposite  to  those  angles? 


A.  That  the  side  ac,  opposite  to  the  smaller  angle  abc, 
in  the  triangle  abc,  is  smaller  than  the  side  AC,  opposite 
to  the  greater  angle  ABC,  in  the  triangle  ABC. 

Q.     How  do  you  prove  this  1 

A.  By  placing  the  triangle  abc  upon  the  triangle 
ABC,  with  the  side  ab  upon  AB  (its  equal),  the  side  be 
will  fall  within  the  angle  ABC,  because  the  angle  abc  is 
smaller  than  the  angle  ABC ;  and  the  ex-tremity  c,  of  the 
line  be,  will  either  fall  without  the  triangle  ABC,  as  you 
see  in  the  figure  before  you,  or  within  it,  or  it  may  also 
fall  upon  the  line  AC  itself. 

Is*.  If  it  falls  without  the  .triangle  ABC,  by  imagining 


GEOMETRY.  49 

the  line  Cc  drawn,  the  triangle  cBC  will  be  isosceles ;  for 
we  have  supposed  the  side  be  equal  to  BC ;  and  because 
i.he  angles  at  the  basis  of  an  isosceles  triangle  are  equal 
(Query  3,  Sect.  II.),  the  angle  z  is  equal  to  the  sum  of 
the  two  angles  x  and  y ;  consequently  greater  than  the 
angle  y  alone;  and  if  the  angle  z  is  greater  than  the 
angle  y,  the  two  angles  z  and  w  together  will  be  greater 
still  than  the  same  angle  y;  therefore,  in  the  triangle 
ACc,  the  angle  AeC  is  greater  than  the  angle  ACc; 
consequently  the  side  AC,  opposite  to  the  greater  angle 
AcC,  must  be  greater  than  the  side  ac,  opposite  to  the 
smaller  angle  ACc.  *  "    ♦  " 

%dly.  If  the  extremity  of  the  line  be  falls  within  the 
triangle  ABC,  the  sum  of  the  two  sides  ac,  be,  must  be 
smaller  than  the  sum  of  the  two 
sides  AC,  BC  (Query  8,  Sect.  II.) ; 
therefore,  by  taking  from  each  of 
these  sums  the  equal  lines  be,  BC, 
respectively,  the  remainder,  AC,  of 
the  greater  sum  (AC-f-BC)  is  greater  than  the  remain- 
der, ac,  of  the  smaller  sum  (ac-\-bc). 

Finally.  If  the  point  c  falls  upon  the  line, AC  itself, 
it  is  evident  that  the  whole  line  AC  must  be  greater  than 
its  part  Ac. 


a 

QUERY  XII. 

If,  in  a  parallelogram,  ACDB, 

you  draw  a  diagonal  CB,  ivhat 

relation    do    the  two    triangles, 

ABC,  CDB,  bear  to  each  other? 

5 


50  GEOMETRY. 

A.  They  are  equal  to  each  other ,  and  the  parallelo- 
gram  is  divided  into  two  equal  parts. 

Q.  How  can  you  prove  this  1 

A.  The  two  triangles,  ABC  and  CDB,  have  the  side 
CB  common ;  and  the  angle  y  is  equal  to  the  angle  w , 
because  y  and  w  are  alternate  angles,  formed  by  the  in- 
tersection of  the  two  parallel  lines  CD,  AB,  by  a  third 
line,  CB;  and  the  angle  x  is  equal  to  the  angle  z,  because 
these  two  angles  are  formed  in  a  similar  manner,  by  the 
parallel  lines  AC,  DB  (Query  10,  Sect.  I.)  :  and  as  the 
triangle  ABC  has  a  side  CB,  and  the  two  adjacent  angles, 
x  and  w,  equal  to  the  same  side  CB,  and  the  two  adjacent 
angles,  z  and  y,  in  the  triangle  CDB,  each  to  each ; 
therefore  these  two  triangles  are  equal  (Query  6,  Sect.  I.), 
and  the  diagonal  CB  divides  the  parallelogram  into  two 
equal  parts. 

Q.  What  other  properties  of  a  parallelogram  can  you 
infer  from  the  one  just  learned? 

1st.  The  opposite  sides  of  a  parallelogram  are  equal; 
that  is,  the  side  CD  is  equal  to  the  side  AB,  and  the  side 
CA  to  the  side  DB;  for  in  the  equal  triangles,  ABC, 
CDB,  the  equal  sides  must  be  opposite  to  the  equal  angles. 
(Conseq.  of  Query  1,  Sect.  II.) 

2dly.  The  opposite  angles  in  a  parallelogram  are 
equal ;  for  in  the  two  equal  triangles,  ABC,  CDB,  the 
same  side,  CB,  is  opposite  to  each  of  the  angles,  at  D  and 
A.     (Conseq.  of  Query  6,  Sect.  I.) 

3dly.  By  one  angle  of  a  parallelogram,  all  four  are 
determined;  for  the  sum  of  the  four  angles  in  a  parallelo- 
gram is  equal  to  four  right  angles ;  because  the  sum  of 
the  three  angles  in  each  of  the  two  triangles,  ABC,  CDB, 
is  equal  to  two  right  angles.  Now,  if  the  angle  at  D,  for 
instance,  is  known,  the  angle  at  A  is  equal  to  it ;  and 
there  remain  but  the  two  angles  ACD  and  ABD,  each 


»> 


GEOMETRY.  £l 

of  which  must  be  equal  to  half  of  what  is  wanting  to 
complete  the  sum  of  the  four  right  angles. 

Q.  If  you  have  a  quadrilateral,  in  which  the  opposite 
sides  are  respectively  equal,  does  it  follow  that  the  figure 
must  be  a  parallelogram  ? 

A.  Yes.  For  if,  in  the  last  figure,  you  have  the  side 
CD  equal  to  the  side  AB,  and  the  side  AC  equal  to  the 
side  BD ;  by  drawing  the  diagonal  BC,  you  have  the 
three  sides  of  the  triangle  ABC,  respectively,  equal  to  the 
three  sides  of  the  triangle  CDB ;  therefore,  these  two 
triangles  are  equal ;  and  the  angle  y,  opposite  to  the  side 
DB,  is  equal  to  the  angle  w,  opposite  to  the  equal  side 
AC ;  and  the  angle  x,  opposite  to  the  side  AB,  is  equal  to 
the  angle  z,  opposite  to  the  equal  side  CD ;  that  is,  the 
alternate  angles,  y  and  w,  x  and  z,  are  respectively  equal : 
therefore  the  side  CD  is  parallel  to  the  side  AB,  and  the 
side  AC  to  the  side  BD,  and  the  figure  is  a  parallelo- 
gram. 

Q.  If,  in  a  quadrilateral,  you  know  but  tioo  sides  to  be 
equal  and  parallel,  what  will  then  be  the  name  of  the 
figure  1 

A.  It  will  still  be  a  parallelogram.  For  if,  in  the  last 
figure,  the  side  CD  is  equal  and  parallel  to  AB,  by  drawing 
the  diagonal  CB,  you  have  the  two  sides,  CB  and  CD,  in 
the  triangle  CDB,  equal  to  the  two  sides,  CB,  AB,  in  the 
triangle  ABC,  each  to  each  ;  and  because  the  side  CD  is 
parallel  to  the  side  AB,  the  included  angle  y  is  equal  to 
the  included  angle  w;  therefore  the  two  triangles  are 
equal  (Query  1,  Sect.  II.),  and  the  side  AC  is  also  equal 
and  parallel  to  the  side  DB,  as  before. 


52  GEOMETRY. 

QUERY    XIII. 

If,  from  one  of  the  vertices  of  a 
rectilinear  figure,  diagonals  .  are 
drawn  to  all  the  other  vertices,  into 
how  many  triangles  will  this  recti- 
linear figure  he  divided? 

A.  Into  as  many  as  the  figure 
has  sides  less  two.    For  it  is  evident,  M  D 

that  if,  from  the  vertex  .A,  for  instance,  you  draw  the  di- 
agonals AF,  AE,  AD,  AC,  to  the  vertices  F,  E,  D,  C, 
each  of  the  two  triangles  AGF,  ABC,  will  need  for  its 
formation  two  sides  of  the  figure,  and.  a  diagonal  ;  but 
then  every  one  remaining  side  of  the  figure  will,  together 
with  two  diagonals,  form  a  triangle ;  therefore  there  will 
be  as  many  triangles  formed,  as  there  are  sides  less  the 
two,  which  are  additionally  employed  in  the  formation  of 
the  two  triangles  AGF,  ABC. 

Q.  And  what  is  the  sum  of  all  the  angles,  'BAG,  AGF, 
GFE,  FED,  EDC,  DCS,  CBA,  equal  to  1 

A.  To  as  many  times  two  right  angles  as  the  figure 
ABCDEFG  has  sides  less  two.  For  as  every  rectilinear 
figure  can  be  divided  into  as  many  triangles  as  there  are 
sides  less  two ;  and  because  the  sum  of  the  three  angles 
in  each  triangle  is.  equal  to  two  rigjit  angles  (Query  13, 
Sect.  I.)  there  will  be  as  many  times  two  right  angles  in 
all  the  angles  of  your  figure,  as  there  are  triangles ;  that 
is,  as  many  as  the  figure  has  sides  less  two. 


SECTION    II. 

PART   II. 


OF  GEOMETRICAL  PROPORTIONS,*  AND  SIMILARITY 
OF  TRIANGLES. 

Whenever  we  compare  two  things  with  regard  to 
their  magnitude,  and  inquire  how  many  times  one  is 
greater  than  the  other,  we  determine  the  ratio  which 
these  two  things  bear  to  each  other.  If,  in  this  way,  we 
find  out  that  the  one  is  two,  three,  four,  &c.  times 
greater  than  the  other,  we  say  that  these  things  are  in  the 
ratio  of  one  to  two,  to  three,  to  four,  fyc. :  e.  g.  If  you 
compare  the  fortunes  of  two  persons,  one  of  whom  is 
worth  $10,000,  and  the  other  $20,000,  you  say,  that 
their  fortunes  are  in  the  ratio  of  one  to  two.  Or  if  you 
compare  two  lines,  one  of  which  is  two,  and  the  other 
six  feet  long,  you  say  of  these  lines,  that  they  are  in  the 

*  It  is  the  design  of  the  author  to  give  here  a  perfectly  element- 
ary theory  of  geometrical  proportions,  and  to  establish  every  prin- 
ciple geometrically,  and  by  simple  induction.  Intending  the 
above  theory  for  those  who  have  not  yet  acquired  the  least  knowl- 
edge of  Algebra,  he  is  not  allowed  to  identify  the  theory  of  pro- 
portions with  that  of  algebraic  equations  (as  it  is  done  by  some 
writers  on  Mathematics),  and  then  to  find  out  the  principles  of  the 
former  by  an  analysis  of  the  latter.  There  are  several  disadvan- 
tages inseparable  from  the  algebraic  method  of  considering  a  ratio 
as  a  fraction,  besides  the  difficulty  of  making  such  a  theory  accessi- 
ble to  beginners.  Neither  can  an  algebraic  demonstration  be 
made  obvious  to  the  eye  like  a  geometrical  one. 
5* 


54  GEOMETRY. 

ratio  of  one  to  three,  because  the  second  line  is  three 
times  as  long  as  the  first. 

It  frequently  occurs,  that  two  things  are  to  each  other 
in  the  same  ratio  in  which  two  others  are ;  we  then  say 
that  these  things  are  in  proportion.  This  is  frequently 
the  case  in  the  fine  arts ;  but  particularly  in  the  science 
of  Geometry,  from  which  these  proportions  are  called 
geometrical.  To  give  an  example :  If  you  draw  a 
house,  you  must  draw  it  according  to  a  certain  scale  ; 
that  is,  you  must  draw  it  one  thousand,  two  thousand, 
three  thousand,  &-c.  times  smaller  than  the  building  itself: 
but  then  you  are  obliged  to  reduce  every  part  of  it  in 
proportion.  If,  for  instance,  you  draw  the  front  of  the 
house  one  thousand  times  smaller  than  the  original,  you 
must  reduce  the  windows,  doors,  and  every  other  part, 
in  the  same  ratio.  If,  on  the  contrary,  the  windows 
were  reduced  two  thousand  times,  whilst  the  doors  and 
other  parts  were  reduced  only  owe  thousand  times^  your 
picture  would  be  out  of  proportion,  because  the  different 
parts  would  be  reduced  by  different  ratios.  In  this  case 
your  picture  would  be  distorted ;  and  would  not  resem- 
ble the  original. 

The  same  is  the  case  with  resemblance,  produced  in 
any  other  kind  of  drawings  ;  but  particularly  in  geomet- 
rical figures. 

C 

Fig.  L       J\  Fig.  II.  Fig.  HI. 


a/\> 


a b    a e 

a c       a c 

I o       b ** 


GEOMETRY.  55 

If  the  two  triangles,  ABC,  abc,  are  to  be  similar  to 
each  other,  it  is  necessary  that  they  should  be  construct- 
ed after  the  same  manner,  and  that  the  side  AC  should 
be  exactly  as  many  times  greater  than  the  side  ac,  as  the 
side  BC  is  greater  than  be,  and  the  side  AB  than  ab.  If 
(Fig.  I.  and  II.)  the  side  AB,  for  instance,  is  twice  as 
great  as  the  .side  ab ;  that  is,  if.  the  side  ab  is  half  of  the 
side  AB ;  the  side  ac  must  also  be  half  of  the  side  AC, 
and  the  side  be  half  of  the  side  BC ;  that  is,  the  three 
sides,  ab,  ac,  be,  of  the  triangle  abc,  must  be  in  propor- 
tion to  the  three  sides,  AB,  AC,  BC,  of  the  triangle  ABC. 
Again,  if  (Fig.  I.  and  III.)  the  side  AB  is  three  times  as 
great  as  the  side  ab ;  that  is,  if  the  side  ab  is  one  third 
of  the  side  AB ;  the  side  ac  must  also  be  one  third  of  the 
side  AC,  and  the  side  be  one  third  of  the  side  BC ;  or 
the  triangles  abc,  ABC,  would  not  be  similar  to  each 
other.  The  same  holds  true  of  all  other  geometrical 
figures,  composed  of  any  number  of  sides.  If  they  are 
similar,  their  sides  are  proportional  to  each  other. 

There  are  different  ways  of  denoting  a  geometrical 
proportion.  Some  mathematicians  express  the  propor- 
tionality of  the  sides,  ab,  ac,  of  the  triangle  abc  (Fig.  II.), 
to  the  sides  AB,  AC,  of  the  triangle  ABC  (Fig.  I.),  in 
the  following  manner : 

AB  :  ab  :  :  AC  :  ac; 
or, 

AB-^-ab  :  :•  AC-^-ac  ; 
and  also 

AB  :  ab  =  AC  :  ac* 
which  is  read  thus  : 

AB  is  to  ab,  as  AC  is  to  ac. 

*  The  first  manner  of  expressing  a  proportion  is  now  in  general 
use  among  the  English  and  French  mathematicians ;  the  second  is 
sometimes  met  with  in  old  English  writers,  and  the  third  way  is 
adopted  in  Germany. 


56  GEOMETRY. 

As  a  proportion  is  nothing  less  than  the  equality  of  two 
ratios,  the  third  way  of  denoting  a  proportion,  in  which 
the  sign  of  equality  is  put  between  the  two  ratios,  seems 
to  be  the  most  natural.  The  reason  why  the  sign  of  di- 
vision (see  Notation  and  Significations),  is  put  between 
the  two  terms,  AB,  ab,  of  a  ratio,  is  obvious ;  for  a  ratio 
points  out  how  many  times  one  term  (the  side  ab)  is  con- 
tained in  the  other  (the  side  AB). 

The  first  and  fourth  terms  of  a  proportion,  together,  are 
called  extremes ;  because  one  of  them  stands  at  the  be- 
ginning,  and  the  other  at  the  end,  of  a  proportion  :  the 
second  and  third  terms,  standing  in  the  middle,  are,  to- 
gether, called  the  means. 

The  following  principles  of  geometrical  proportions 
ought  to  be  well  understood  and  remembered  r 

1st.  It  is  important  to  observe,  that  in  every  geometrical 
proportion  the  two  ratios  may  be  inverted;  that  is,  in- 
stead of  saying, 

AB  :  ab  =  AC  :  ac, 
you  may  say, 

ab  :  AB  =  ac  :  AC ; 

for,  the  order  of  terms  being  changed  in  both  ratios, 
they  are  still  equal  to  one  another  ;  but,  leaving  one  ratio 
unaltered,  if  you  change  the  order  of  terms  in  the  other, 
the  proportion  will  be  destroyed.     You  cannot  say, 

ab  :  AB  =  AC  :  ac  ; 

for  the  smaller  side,  ab,  is  contained  twice  in  the  greater 
side,  AB  (Fig.  I.  and  II.) ;  but  the  greater  side,  AC,  is 
not  contained  once  in  the  smaller  side,  ac. 

2d.  Another  remarkable  property  of  geometrical  pro- 
portions is,  that  you  may  change  the  order  of  the  mcanr,, 
or  extremes,  without  destroying  the  proportion.  Thus 
you  may  change  the  proportion 


Geometry.  57 

AB  :  ab  =  AC  :  ac      .....       (I.) 

into 

AB  :  AC  =  ab  :  ac  .     .     .     .     .     .    (II.) 

or  by  changing  the  extremes  into 

ac  :  ab  =  AC  :  AB       .....  (III.) 

The  reason  why  you  have  a  right  to  do  this,  is  easily 
comprehended.  If,  in  the  first  proportion,  the  side  AB 
is  as  many  times  greater  than  ab,  as  AC  is  greater  than 
ac,  the  ratio  of  AB  to  AC  will  be  the  same  as  that  of 
ab  to  ac.  In  Fig.  I.  and  II.  (page  54),  we  have  ab 
equal  to  one  half  of  AB  ;  consequently  ac  is  also  equal 
to  one  half  of  AC ;  and,  therefore,  let  the  ratio  of  the 
two  lines,  AB  to  AC,  be  whatever  it  may,  their  halves, 
ab  and  ac,  must  be  in  the  same  ratio.  To  give  another  ex- 
ample :  If  A's  garden  is  five  times  greater  than  B's,  half 
of  A's  garden  is  also  five  times  greater  than  half  of  B's 
garden.  The  second  proportion  (II.)  would  still  be  correct, 
if,  as  in  Fig.  I.  and  III.,  the  sides  AB,  AC,  were  three 
times  as  great  as  the  sides  ab,  ac ;  for  then  the  thirds 
of  AB  and  AC  would  still  be  in  the  same  proportion  as  the 
whole  lines  AB  and  AC.  Nothing  can  now  be  easier 
than  to  extend  this  mode  of  reasoning,  and  show  the 
generality  of  the  principle  here  advanced.  The  correct- 
ness of  the  third  proportion  might  be  proved  precisely  in 
the  same  manner  as  that  of  the  second ;  for  the  third 
proportion  (III.)  differs  from  the  second  (II.)  only  in  the 
order  in  which  the  two  ratios  are  placed ;  and  of  two 
equal  things,  it  does  not  matter  which  you  put  first.  The 
correctness  of  the  second  proportion  proves,  therefore, 
that  of  the  third  proportion. 

3d.  If  you  have  two  geometrical  proportions,  which 
have  one  ratio  common,  the  two  remaining  ratios  will, 
again,  make  a  proportion ;  for  if  two  ratios  are  equal  to 
the  same  ratio,  they  must  be  equal  to  each  other.  (See 
Axioms,  Truth  I.)     If  you  have  the  two  proportions 


58  GEOMETRY. 

AB  :  ab  =  AC  :  ac 
AB  :  ab  —  BC  :  be 

you  will  also  have  the  proportion 

AC  :  ac  =  BC  :  6c. 
For  an  illustration  of  this  principle,  we  may  take  the  two 
triangles  ABC,  abc  (Fig.  I.  and  II.)  :  If  the  sides  AB 
and  ab  are  in  proportion  to  the  sides  AC  and  ac,  and 
also  in  proportion  to  the  sides  BC  and  be,  the  three  sides 
of  the  triangle  ABC  will  be  in  proportion  to  the  three 
sides  of  the  triangle  abc;  therefore,  any  two  sides  of  the 
first  triangle  will  be  in  proportion  to  the  two  correspond- 
ing sides  of  the  other  triangle. 

4th.  Another  important  principle  of  geometrical  pro- 
portions is  this:  If  you  have  several  geometrical  propor- 
tions, of  which  the  second  has  a  ratio  common  with  the 
first,  the  third  a  ratio  common  with  the  second,  the  fourth 
u  ratio-common  with  the  third,  and  so  on;  the  sum,  of  all 
thejiftt  terms  of  these  proportions  will  bear  the  same  ratio 
to  the  sum  of  all  the  second  terms,  which  the  sum  of  all  the 
third  terms  docs  to  the  sum  of  all  the  fourth  terms,  that 
ti>,  the  sums  will  again  make  a  proportion. 

To  prove  this,  we  will,  in  the  first  place,  consider  the 
simplest  case  ;  that  of  two  proportions  only  ;  and,  the 
easier  to  comprehend  it,  take  the  same  two  proportions 
wnich  we  have  just  had  under  consideration,  viz  : 

ac  :  AC  =  ab  :  AB 

ab  :  AB  =  be  :  BC. 
We  know,  from  the  two  triangles,  ABC  and  abc  (Fig.  I. 
and  II.),  that,  in  the  first  proportion,  ac  is  half  of  AC; 
consequently  ab  is  also  half  of  AB,  and,  in  the  second 
proportion,  be  is  also  half  of  BC.  Thus,  each  of  the  two 
first  terms,  ab,  ac,  is  half  of  its  second  term  ;  and  conse- 
quently each  of  the  third  terms,  be,  ab,  is  half  of  its  cor- 
responding fourth  term ;    therefore,   adding  ab  and  ac 


GEOMETRY.  59 

together,  their  sum  will  be  one  half  of  the  sum  of  AB 
and  AC ;  and  so  will  be  and  ab,  be,  together,  one  half  of 
the  sum  of  BC  and  AB.  For  the  sake  of  illustration, 
you   may  measure  off  the  length  of  ab  and  ac,  upon 


a 


B 


the  line  be,  and  the  length  of  AB  and  AC  on  another 
line  BC  ;  and  you  will  find  that  the  line  be  is  exactly  one 
half  of  the  line  BC.  For  the  line  be,  composed  of  two 
parts,  ab,  ac,  each  measuring  exactly  one  half  of  the 
corresponding  two  parts,  AB,  AC,  of  which  the  line  BC 
is  composed,  must  evidently  be  one  half  of  the  whole  line 
BC.     In  the  same  way  you  may  convince  yourself  that 


cv 


A  B  C 

the  line  ac,  composed  of  the  two  parts,  ab  and  be,  meas- 
ures one  half  of  the  second  line  AC,  composed  of  the  two 
parts' AB  and  BC  :  and  therefore  you  will  have 

ab-\-ac  :  AB  -f  AC  =  be  -f  ab  :  BC  -f  AB.* 
Although,  in  our  example,  we  have  chosen  a  proportion 
in  which  the  first  and  third  terms  are  exactly  one  half 
of  the  second  and  fourth  terms,  yet  it  is  easy  to  perceive, 
that  the  same  course  of  reasoning  will  apply  to  any  other 
two  proportions.  Thus,  if  the  first  terms  in  the  above 
proportions  were  one  third,  or  one  fourth,  or  one  fifth, 

*  The  lines  over  ab-\-ac,  AB-f-AC,  &c,  mark  that  ab-\-ac, 
AB-|- AC,  &c,  are  but  single  lines  composed  of  the  two  parts,  ab, 
ac,  and  AB,  AC. 


60  GEOMETRY. 

&*c,  of  the  corresponding  second  terms,  the  sum  of  all 
the  first  terms  would  also  be  one  third,  or  one  fourth,  or 
one  fifth,  &,c,  of  the  sum  of  all  the  second  terms ;  and 
the  same  would  be  the  case  with  regard  to  the  sum  of  the 
third  and  fourth  terms.  It  is  also  evident,  that  our  prin- 
ciple would  still  hold  true,  if,  instead  of  two  proportions, 
we  had  three,  four,  or  more  proportions  given,  of  which 
two  and  two  have  a  common  ratio.  If,  for  instance,  we 
had  the  three  proportions 

ac  :  AC  =  ab  :  AB 

ab  :  AB  =  be  :  BC 

be  :  BC  =  ac  :  AC, 
we  should,  according  to  our  principle,  have 


be  -f-  ab  -f  ac  :  BC  -f-  AB  -f  AC  —  ac  -\-  be  -f  ab 
:  AC  +  BC-f-AB. 
Each  of  the  three  lines,  be,  ab,  ac,  would  be  one  half  of 
its  corresponding  second  term ;  and  in  the  same  way 
would  each  of  the  three  lines,  ac,  be,  ab,  be  one  half  of 
its  corresponding  fourth  term ;  and,  therefore,  the  sum  of 
the  three  lines,  be,  ab,  ac,  or,  which  is  the  same,  a  line 
as  great  as  the  three  lines,  be,  ab,  ac,  together,  would  be 
one  half  of  the  sum  of  the  three  lines,  BC,  AB,  AC,  or  a 
line  as  great  as  the  three  lines,  BC,  AB,  AC,  together  ; 
and  the  same  would  be  the  case  with  the  sum  of  the  third 
and  fourth  terms.  And  in  like  manner  can  this  principle 
be  extended  to  four,  five,  six,  and  more  proportions. 

5th.  Another  principle,  which  it  is  important  to  recol- 
lect, is,  that  by  adding  the  second  term  of  a  proportion 
once,  or  any  number  of  times,  to  the  first  term,  and  the 
fourth  term  the  same  number  of  times  to  the  third  term, 
you  will  still  have  a  proportion.  To  give  an  example : 
In  the  proportion 


•*■ 


GEOMETRY.  Ql 

AB  :  ab  =  AC  :  ac, 
let  there  be  added  the  second  term  ab,  in  the  first  place, 
once  to  the  first  term  AB ;  and  the  fourth  term  ac  also 
once  to  the   third   term. AC.     Our   proportion   is   then 
changed  into 

AB  +  a0  '■  a0  —  AC  -f-  ac  :  ac, 
in  which  the  first  term,  AB  -f-  ab,  instead  of  being  only 
twice  as  great  as  ab;  is  now,  by  the  addition  of  the  term 
ab  itself,  three  times  as  .great  as  ab  ;  and  for  the  same 
reason  is  AC  -f-  ac  three  times  as  gr,eat  as  ac:  The  two 
new  ratios,  "•■'*,  •   -     ■ 


AB -\-ab  :  ab,and 
AC  -f-  ac  :  ar,  '  "• 

are  therefore  equal,  and  consequently  make  a  proportion. 
The  same  would  be-  the  case,*  if,  instead  of  adding  the 
second  and  fourth  terms  once,  you  would  add  them  twice 
respectively  to.  the  first  and  third  terms  ;  with  the  only 
difference,  that  the  first  term,  AB  -{-Hab,  would  then  be 
four  times  as  great  as  the  second  term  ab.  A  similar 
change  would  lake  place  with  regard  to  the  third-  term, 
AC  -}-  2ac,  which  would  then  be  four  time's  as  great  as 
the  term  ac ;  and  you  would  have  the  proportion 

AB-\-2a6  \.ab  —  AC  -f  2ac  :  ac.  i 
If  the  second  term  were  added .  three  times  to  the  first 
term,  the  first  term,  AB  -f-  'Sab,  would  be  five  times-  as 
great  as  ah;  and  the  third  term,  AC  -j-  Sac,  would  also 
be  five  times. as  great  as  ac  i  and  so  on. 

In  precisely  the  same  manner  you  may  prove  that,  by 
adding  ilit  first  term  q nee,  or  any  number  of  times,  to  the 
second  term,  and  the  third  term  the  same,  number  of  times 
to  the  fourth  term,  the  result  will  still  be  a  proportion. 
Thus,  our  proportion 
6 


62  GEOMETRY. 

AB  :  ab  —  AC  :  ac, 

may  be  changed  into 


AB  :  a6  +  AB  =  AC  rac-j-AC, 
or  into  \  ~ 


AB  :  a6-j-2AB  — AC  :ac-|-2AC,  &c. 

It  is  also  evident  that  the  same  principle  will  hold  of  any 
other  geometrical  proportion.* 

Gth.  For  the  same  reason  that  the  second  term  of  a 
geometrical  proportion  may  be  added  once  or  any  number 
of  titties'  to  the  first  term,  and  the  fourth  term  the  same 
number  of  times  to  the  third  term,  without  destroying  the 
proportion;  the  second  term  may  also  be  subtracted  once 
or  any  number  of  times  from  the  first  term,  provided  the 
fourth  term,  is  the  same  number  of  times  subtracted  from 
the  third  term,  and  the  result  will  still  be  a  proportion. 

If,  in  the  geometrical  proportion 

AB  :  ab  —  AC  :  ac , 
the  first  term  (AB)  is  twice  as  great  a.s  ab,  and  AC  twice 
as  great  as  «c,  we  shall, .by  subtracting  ab  from  AB,  and 
ac  from  AC,  make  the  two  terms  jn  each  ratio  equal ; 
and  we  shall  have  a  new  proportion, 

A3  —  ab:ab  =2  AC  —  ac  :  ac. 

*  The  teacher  had  better  show  this  to  the  pupil,  particularly 
as  the  above  mode  of  demonstrating  this  principle  admits  of  an 
ocular  demonstration  by  measurements.  For  if  the  "teacher  uses 
lines  for  the  terms  of  his  proportions,  and  not  abstract  numbers, 
which  are  always  more  difficult  to  be  comprehended,  he  can  actu- 
ally perform,  these  additions,  by  extending  the  Hne  AB,  for 
instance,  to  once  or  twice  the  length  of  the  line  ab,  and  then  show, 
by  measuring  these  lines,  that  the  fjrst  term  is  really  as  many 
times  greater  than  the  second  term,  as  th^e  third"  terra  is  greater 
than  the  fourth  term.  In  this  manner  the  demonstrations  will  not 
only  be  perfectly  geometrical,  but  also  have  the  advantage  of  the 
'jvhictive  method. 


GEOMETRY.  63 

If  AB  were  three  times  as  great  as  ab,  AC  would,  of 
sourse,  be  three  times  as  great  as  ac  ;  and  therefore,  by 
subtracting  ab  from  AB,  and  ac  from  AC,  the  first  term 
(AB  —  ab),  in  the  last  proportion,  would  be  twice  as 
great  as  ab;  and  for  the  same  reason  would  AC  —  ac,  be 
twice  as  great  as  ac.  In  the  same  manner  may  this 
principle "  be  applied  to  ever/  other.,  geometrical  propor- 
tion ;  and  it  may  also  be  proved,  that,  by  subtracting  the 
first  term  of  a  geometrical  proportion  once  or  any  number 
of  times  from  the  second  term^  and  the  third  term  the  same 
number  of  times- from  the  fourth  term^  the  proportion  icill 
not  be  destroyed.:      ■  . 

7th..  If  all  the  ■  terms  of  a  geometrical  proportion  are 
multiplied  or  -divided  by  the  same  number,  the  proportion 
remains  the  same. 

For  an  example,  we  will  again  take  the  proportion 
AB  :  ab  3=  AC  :  ac, 
in  which  ab  is  half  of  AB,  and  ac  half  of  AC.  Then  it 
is  evident,  that  a  line,  which  is,  for  instance,  ten  times 
as  long  as  ab,  that  is,  a  line  which  contains  the  line  ab 
ten  times,  is  still  half  of  a  line  winch  contains  the  line 
AB  ten  times;  and  in  like  manner  is  a  line  ten  times  as 
long  as  ac  still  half  of  a  line  ten  times  as  long  as  AC  ; 
consequently  the  proportion 

10  AB  :  10a&=:10AC  :  10  ac 
is. the  same  as 

AB  :  ab  —  AC  :  ac, 
because  in  both  proportions  the  first  term  in  each  ratio  is 
double  of  the  second  term. 

Neither  would  our  proportion  change,  if,  instead  of 
multiplying  each  term  by  10,  we  were  to  multiply  it  by  2, 
by  3,  by  4,  &c,  or  even  by  fractions;  for  the  reasoning 
would,  in  every  one  of  these  cases,  be  precisely  the  same 
as  in  the  case  of  our  multiplying  by  ten. 


64  GEOMETRY. 

It  is  also  easy  to  apply  the  same  principle  to  any  other 
geometrical  proportion. 

If,  instead  of  multiplying  each  term  of  the  proportion 
AB  :  ah  sen  AC  :■  ac, 
we  divide  it  by  ten,  it  is  evident  that  the  tenth  part  of 
the  line  ah  will  still  be  half  of  the  tenth  part  of  the  line 
AB ;  and  so  will  the  tenth  part  of  the  line  ac  be  half 
of  the  tenth  part  of  the  line  AC  ;  consequently  the  pro- 
portion 

TVAB  :  T^ah  —  T\AC  :  T\ac 
is  still  the  same  as 

AB  :  ah  =  AC  :  ac ; 
and  the  same  reasoning  may  be  applied  to  the  division 
by  any  other  number,  and  to  any  other  geometrical  pro- 
portion. 

8th.  If  three  terms  of  a  proportion  he  given,  the  fourth 
term  can  easily  he  found.  Let  there  be  the  three  terms 
of  a  proportion, 

AB:«S  =  AC: 
to  which  the  fourth  term  is  wanting.  Then,  by  knowing 
how  many  times  the  line  ab  is  smaller  than  the  line  AB, 
or.,  which  is  the  same,  whatever  part  of  the  line  AB  the 
line  ah  is,  you  can  easily  take  the  same  part  of  the  line 
AC,  which  will  be  the  fourth  term  of  your  proportion. 
If  you  know,  for  instance,  that  the  line  ab  is  one  lialf  of 
the  line  AB,  you  would  at  once  conclude,  that  the  re- 
quired fourth  term  in  your  proportion  must  be  one  half 
of  the  line  AC  :  this  is,  as  we  know,  really  the  case  with 
our  proportion,  where  the  fourth  term  ac,  which  we  sup- 
posed here  to  be  unknown,  is  really  one  half  of  AC.  If 
ah  were  one  third  of  AB,  you  would  conclude  that  your 
fourth  term  must  be  one  third  of  AC  ;  and  so  on.  If, 
instead   of  the   fourth   term,  another,  for   instance   the 


GEOMETRY.  65 

second  term,  were  unknown,  you  could  find  it  in  a  man- 
ner similar  to  the  one  just  given.  For,  one  ratio  being 
expressed,  you  will  always  know  the  relation  which  £fl 
term  you  are  to  find  must  bear  to  the  term  with  which  it 
is  to  form  a  ratio.  .  - 

9th.  Geometrical  proportions  are  also  frequently  made 
use  of  in  common-  Arithmetic ,  and  in  Algebra.  You  can 
say  of  the  two  numbers  3  and  6,  that  they  are  in  propor- 
tion to  the  numbers  4  and  8  ;  because  3  are  as  many 
times  contained  in  6,  as  4  in  8,  which  may  be  expressed 

thus :         "  • 

3  :6  =  4:  :8. 

For  this  reason,  if  four  lines  are  in  a  geometrical  pro- 
portion, their  length,  expressed  in  numbers  of  rods,  feet, 
&c,  will  be  in  the  same  proportion. 

10th.  It  is  to  be  remarked,  that  in  every  geometrical 
proportion,  expressed  in  numbers*  the  product  obtained 
by  multiplying  the  two  mean  terms  together,  is  equal 
to  the  product  obtained  by  multiplying  the  two  extreme 
terms.  In  the  above  proportion,  3:6=4:8,  for  in- 
stance, we  have,  3  times  8  equal  to  6  times  4.  For,  the 
first  of  our  extreme  terms,  3,  being  exactly  asjnany  times 
smaller  than  the  first  of  our  mean  terms,  6,  as  the  second 
of  our  extreme  terms,  8,  is  greater  than  the  second  of 
our  mean  terms,  4  (namely,  twice)  ;  what  the  multiplier 
3,  in  the  one  case,  is  smaller  than  the  multiplier  6,  in  the 
other,  is  made  up  by  the  multiplicand  8,  which  is  as  many 
times  greater  than  the  multiplicand  4,  as  the  multiplier  3 
is  smaller  than  the  multiplier  6 ;  and  in  a  similar  manner 


*  For  wc  cannot  multiply  lines  together,  but  merely  the  abstract 
numbers,  which  express  their  relative  length. 

6*  -r 


66  GEOMETRY. 

we  could  prove  the  same  of  any  other  geometrical  pro- 
portion. To  give  but  one  more  example :  In  the  pro- 
portion 

2:4r=3:_6, 
we  have,  again,  twice  6  equal  to  4  times  3  ;  because  the 
first  multiplier,  2,  is  exactly  as  many  times  smaller  than  the 
second  multiplier,  4,  as  the  first  multiplicand,  6,  is  greater 
than  the  second  multiplicand  3  (namely,  twice).* 

If  both  ratios  of  our  proportion  were  inverted,  as,  for 
instance,  4  :  2  —  6  :  3,  our  principle  would  still  prove  to  be 
correct.  For  we  have  again  4  times  3  equal  to  twice  6. 
The  only  difference  consists  in  the  mean  terms  having 
now  become  the  extreme  terms,  and  vice  versa.  If  we 
change  the  order  of  the  means  and  extremes^  their  pro- 
ducts refnain  still  the  same.  For  3  times  8  are  the  same 
as  8  times  3  ,v  ^nd  6  times  2  the  same  as. twice  6. 

•When,  in  a  geometrical  proportion,  the  two  mean 
terms  are  equal  to  one  another,  either  of  t-htem  is  called  a 
mean  proportional  between  the  two  extremes.  Thus,  in 
the  proportion,     .     ' 

4:6  =  6:9, 

6  is  a  mean  proportional  between  4  and  9. 

What    you    have   learned  of  geometrical    proportions 
will  enable  you-  to  understand  every  principle  in  plane 
Geometry  ;  we  will  therefore  continue  our  inquiries  into 
the  principles  of  geometrical  figures. 
, 

*  The  teacher  may  illustrate  this  principle  by  a  balance  ;  show- 
ing that  2  weights,  of  6  pounds  each,  are  in  equilibrium  with  4 
weights  of  3  pounds  each.  The  weights  in  this  example,  6  pounds 
and  3  pounds,  are  the  multiplicands,  and  their  number  2  and  4  are 
the  respective  multipliers. 


GEOMETRY.  07 


QUERY  XIV. 

If  you  divide  one  side,  AB~ 
of  a  triangle,  ABC,  into  any 
number  of  equal  parts,  for 
instance  four,  and  then,  from 
the  points  of  division  D,  F,  H, 
draw  the  lines  DE,  FG,  JIK, 
parallel  to  the  side  AC,  ivhat 
remark  can  you  make  with  regard  to  the  other  side  BC? 

A.  That  the  other  side,  BC,  is  divided  into  as  many 
equal  parts  as  the  side  AB. 

Q.  How  can  you  prove  this? 

A.  By  drawing-  the  lines  DL,  FM,  HN,  parallel  to  the 
side  BC,  the  triangles,  BDE>  DFL,  FHM,  ilAN,  are  all 
equal  to  one  another.  For,  comparing,  in  the  first  place, 
the  two  triangles,  BDE,  DFL,  we  see  that  the  side  BD 
is  equal  to  DF  (beeause  we  have  divided  the  line  AB 
into  equal  parts)  ;  and  the  angle  x  is  equal  to  the  angle  z, 
because  these- angles  are  formed  by  the  two  parallels  DL 
and  BC,  being  intersected  by  the  straight  line  AB  (Query 
10,  Sect  I.) ;  and  the  angle  y  is  equal  to  the  angle  iv, 
because  y  and  w  are  formed,  in  a  similar  manner,  by  the 
two  parallels,  DE,  FG,  being  intersected  by  the  same 
straight  line  AB  :  consequently  we  have  one  side,  DB, 
and  the  two  adjacent  angles  x  and  y,  in  the  triangle  BDE, 
equal  to  one  side  DF,  and  the  two  adjacent  angles,  *  and 
w,  in  the  triangle  DFL  ;  therefore  these  two  triangles  are 
equal  to  each  other  (duery  6,  Sect  I.) ;  and  the  side  DL, 
opposite  to  the  angle  w,  in  the  triangle  DFL,  is  equal  to 
the  side  BE,  opposite  to  the  equal  angle  y,  in  the  triangle 
BDE ;  and  in  the  same  manner  it  can  be  proved,  that 
FM  and  HN,  are  also  equal  to  BE.  Now,  each  of  the 
quadrilaterals,  DELG,  FGMK,  HKNC,  is  a  parallels 


OS  GEOMETRY. 

gram  (because  the  opposite  sides  are  parallel) ;  and  as 
the  opposite  sides  of  a  parallelogram  are  equal,  DL  must 
be  equal  to  EG,  FM  to  G.K,  and  HN  to  KC.  But  each 
of  the  lines  DL,  FM,  HN,  is  equal  to  BE;  therefore, 
each  of  the  lines  EG,  GK,  KC,  must  also  be  equal  to 
BE  ■  consequently  the  line.  BC  is  divided  into  the  same 
number  of  equal  parts  as.  the  line  AB. 

Q.  Could  you  prove  the  same  principle  in  the  case 
where  the  line  AB  is  divided  into  five,  six,  or  more  equal 
parts  1  ■ 


QUERY  XV. 

If,  in  a  triangle j  ARC, .you  draw  a 
line,DE,  parallel  to  one  of  the  sides, 
say  AC;  what  relation  do  the  parts 
BD,DA;  BE,  EC,  into  which  the 
sides  AB  and  BC  qre  divided,  bear 
to  each  other,  and  to  the  whole  of  the 
sides  AB,  BC  ? 

A.    The  upper  parts,  BD  and  BE, 
as  well  as  the  loivcr  parts,  DA  and  EC,  arc  in  the  same 
ratio,  in  ichich  the  whole  sides  AB,  BC,  themselves  are. 

Q.  Why? 

A.  Because  you  can  imagine  the  side  AB  to  be  suc- 
cessively divided  into  smaller  aiid  smaller  parts,  until  one 
of  the  points  of  division  shall  have  fallen  upon  the  point 
D  :  then,  by  drawing,  through  all  the  points  of  division, 
parallel  lines  to  the  side  AC,  the  side  BC  will  be 
divided  into  as  many  equal  parts*  as  the  side  BA  (last 
Query)  ;  and  as  the  line  DE  itself  will  be  one  of  these 
parallels,  BE  will  have  as  many  of  these  parts  marked  as 
BD ;  and  EC  as  many  as  DA  :  and  therefore  the  ratio  of 


GEOMETRY.  69 

the  whole  of  the  line  BA  to  the  whole  of  the  line  BC,  must 
be  the  same  as  that  of  BD  to  BE,  or  DA  to  EC. 

Q.  How  can  you  express  these  proportions  in  writing  ? 
A.  BA  :  BC  =  BD  :  BE 

B A  :  BC  =  DA  :  EC ; 
consequently,  also, 

BD:BE  =  DA:EC 

(3d  principle  of  proportion). 

Q.  Is  the  reverse  of  the  same  principle  also  true  ?  that 
is,  must  the  line  DE  be  parallel  to  AC,  when  the  parts 
BD  and  BE,  and  DA  and  EC,  are  proportional  to  each 
other,  or  to  the  whole  of  the  sides  BA,  BC  ? 

A>  Yes.  For  you  need  only  imagine  the  side  BA  to 
be  again  successively  divided  into  smaller  and  smaller 
parts,  until  one  of  the  points  of  division  shall  have  fallen 
upon  D.  Then,  it  is  evident  that,  by  drawing,  as  before, 
through  the  points  of  division,  parallels  to  the  side  AC, 
DE  itself  must  be  one  of  them,  if  BE  shall  again  have  as 
many  of  these  parts  marked  as  BD,  and  EC  as  many  as 
DA  ;  for  only  in  this  case  can  BE,  BD,  and  EC,  DA,  be 
proportional  to  each  other,  and  to  the- whole  of  the  sides 
BC  and  BA. 


Remark.  It  has  already  been  stated  (page  55),  that  two  geomet- 
rical figures  Cannot  be  similar  to  each  other,  unless  they  are  con- 
structed after  the  same  manner,  and  have  their  sides  proportional. 
We  will  now  give  the  strictly  geometrical  definition  of  the  same 
principle  for  rectilinear  figures.  . 

In  order  that  two  rectilinear  figures  may  be  similar  to  each 
other,  it  is  necessary, 

1st,  That  both  figures  should  be  composed  of  the  same  number 
of  sides /*  • 

*  This  will,  of  course,  always  be  the  case  in  triangles. 


70  GEOMETRY. 

2dly,  That  the  angles  in  one  figure  should  be  equal  to  the 
angles  in  the  other,  each  to  each  ; 

3dly,  That  these  angles  should  follow  each  other  in  precisely 
the  same  order  in  both  figures  ;   and, 

4thly,  That  the  sides,  ivhich  include  the  equal  angles  in  both 
figures  (and  which  are  therefore  called  the  corresponding  or 
homologous  sides*),  should  be  in  a  geometrical  proportion. 

QUERY  XVI. 

If,  in  a  triangle,  ABC,  you  draw 
a  line,  DE,  parallel  to  one  of  the 
sides,  say  AB,  tvhat  relation  docs 
the  triangle  DEC,  which  is  cut  off , 
bear  to  the  wliole  of  the  triangle 
ABC? 

A.  The  triangle  J) EC  is  simi- 
lar to  the  triangle  ABC. 

Q    Why?  .,../'.% 

A.  Because  tli e  three  angles,  x,  y,  z,  of  the  triangle 
DEC,  are  equal  to  the  three  angles,  w,  y,  t,  of  the  triangle 
ABC,  each  to  each ;  for  the  angles  x  and  z  are  respec- 
tively equal  to  the  angles  w,  t;  because  the  line  DE  is 
parallel  to  AB  (Query  10,  Sect.  I.).  This  satisfies  the 
three  first  conditions  of  similarity.  Moreover,  we  have 
the  proportion  CD  :  CE  rz:  CA  :  CB  (preceding  Query), 
and  by  drawing  DH  parallel  to  the  side  CB,  also  the  pro- 
portion CD  :  BH  (or  ED)  r=  AC  :  AB ;  therefore,  the 
three  sides  of  the  triangle  DEC  are  proportional  to  the 
three  sides  of  the  triangle  ABC,  which  is  the  fourth  con- 
dition of  similarity  :  consequently  these  two  triangles  are 
similar  to  each  other. 


*  In  triangles,  the  corresponding  sides  are  those  which  are  oppo- 
site to  the  equal  angles. 


GEOMETRY.  71 


QUERY  XVII. 

If  the  three  angles  in  one 
triangle  are  equal  to  the 
three  angles  in  another  tri- 
angle, each  to  each,  ivhat 
relation  do  these  triangles 
bear  to  each  other  ? 

A.  They  arc  similar. 

Q.  How  can  you  prove'  it  1 

A.  By  applying  the.  triangle  abc  to  the  triangle  ABC, 
the  angle  at  c  will  coincide  with  the  angle  at  C,  and  the 
side  ca  will  fall  upon  CA,  and  cb  upon  CB  ;  and  as  the 
angles  at  a  and  b,  in  the  triangle  abcl  are  respectively 
equal  to  the  angles  at  A  and  B,  in  .the  triangle  ABC,  the 
side  ab  will  fall  parallel  to  the  side  AB  (Query  8,  Sect. 
I.)  and  we  shall  have  the  same  case  as  in  the  preceding 
Query:  consequently,  the  triangles  abc  and  ABC  will 
be  similar  to  each  other. 

Q.  Supposing  you  have  a  triangle,  of  which  you  know 
only  two  angles,  respectively,  equal  to  two  angles  in  another 
triangle,  whaf  can  you  infer  with  regard  to   these   two 


A.  -That  tliey  must  still  be  similar  to  each  other.  For 
two  angles  of  a  triangle  always.,  determine  the  third  one 
(page  34,  2d.). 

QUERY  XVIII. 

If  you  have  two  triangles,  abc,  ABC  (see  the  last 
figure),  and  only  know  that  one  angle  at  c,  in  the  one,  is 
equal  to  one  angle  at  C  in  the  other,,  but  that  the  sides, 
which  include  that  angle  in  both  triangles,  are  in  a  geo- 
metrical proportion,  what  inference  can  you  draw  from  it  1 

A.    That  these   triangles   are  again   similar  to  each 


72 


GEOMETRY. 


other.  For  if  you  imagine  the  triangle  abc  placed,  as 
before,  upon  the  triangle  ABC,  the  angle  at  c  will  again 
coincide  with  the  angle  at  C,  and  the  side  ca  will  fall 
upon  CA,  and  cb  upon  CB ;  and  as  ca  and  cb  are  pro- 
portional to  CA  and  CB,  the  side  ab  will  fall  parallel  to 
the  side  AB  (Query  15,  Sect.  II.);  and- we  shall  once 
more  have  the  same  case  as  in  Query  16,  Sect:  II. ;  con- 
sequently the  triangle  abc  will  be  similar  to  the  triang\e 
ABC 


QUERY   XIX. 

L°.t  us  now  consider  the  case,  where  all  the  angles  of 
two  triangles  arc  unknown-,  but  the  three 'sides -of  the 
one  (ire  in  prdportion  to  the  .  three  sides  of  the  other ; 
what  relation  will  these  tfiahglcs  hear  io  each  other? 

A.    They  will  still  be  similar  to  each  other  ? 

Q.  How  can  you  prove  it  ?  J 

A.  Let  us  suppose,  for  instance, 
that  the  three  sides  of  the  triangle 
ab'  are  in  proportion  to  the  three 
sides  of  the'  triangle  ABC  ;    that 
is,  iet  us  have  the  proportions 
ac  :  ab  =  AC  :  AB 
ac  :  cb  =  AC  :  CB.     . 
Then  make  CD  equal  to   ca,  and 
draw    through    the    point    D    the 
line  DE,  paralfel"  to  AB ;  and  the 
triangle   CDE   will  be  similar  to 
the    triangle     CAB     (Query    16, 
Sect.  II.),  and  we  shall  "have  the 
proportions 

DC  :  I>E  =  AC  :  AB 

DC  :  CE  =  AC  :  CB, 

in  which  the  two  ratios,  AC  :  AB,  and  AC  :  CB,  are  he 

same  as  in  the  first  two  proportions ;  consequently,  eom- 


GEOMETRY.  73 

paring  these  two  proportions  with   the   two  preceding 
ones,  we  shall  have 

DC  :DE  =  ac:ab 

DC  :  CE  =  ac  :  cb 
(see  theory  of  proportions,  principle  3d). 

Now,  as  I  have  made  DC  equal  to  ac,  I  can  write  ac 
instead  of  DC,  in  the  two  last  proportions ;  and  they  will 
then  become 

ac  :  DE  =  ac  :  ab, 

ac  :  CE  =  ac  :  cb. 
The  upper  one  expresses,  that  the  line  DE  is  as  many 
times  greater  than  ac,  as  the  line  ab  is  greater  than  the 
same  line  ac  (Definition  of  geometrical  proportions)  ; 
consequently  the  line  DE  is  equal  to  the  line  ab.  In 
like  manner  does  the  lower  one  express,  that  CE  is  as 
many  times  greater  than  ac,  as  cb  is  greater  than  the 
same  line  ac ;  consequently  CE  is  also  equal  to  cb  ;  and 
the  three  sides  of  the  triangle  DEC,  are  equal  to  the  three 
sides  of  the  triangle  abc,  each  to  each ;  therefore  these 
two  triangles  are  equal  to  one  another  (Query  4,  Sect. 
II.)  ;  and  as  the  triangle  DEC  is  similar  to  the  triangle 
4.BC  the  triangle  abc  will  also  be  similar  to  it. 


Q.  Will  you  now  briefly  state  the  different  cases,  in 
which  two  triangles  are  similar  to  one  another  1 

A.  1st.  When  the  three  angles  in  one  triangle  are 
equal  to  the  three  angles  in  another,  each  to  each ;  and 
also  when  two  angles  in  one  triangle  are  equal  to  two 
angles  in  another,  each  to  each ;  because  then  the  third 
angle  in  the  one  is  also  equal  to  the  third  angle  in  the 
other. 

2dly.  When  an  angle  in  one  triangle  is  equal  to  an 
angle  in  another,  and  the  two  sides  which  include  that 
7 


74  GEOMETRY. 

angle  in  the  one  triangle,  are  in  proportion  to  the  tioo 
sides  which  include  that  angle  in  the  other  triangle. 

3dly.   When  the  three  sides  of  one  triangle  are  in  pro- 
portion to  the  three  sides  of  another. 

QUERY  XX. 

If  you  have  a  right-an- 
gled triangle,  ABC,  and 
from  the  vertex  A,  of  the 
riglit  angle,  let  fall  a  per- 
pendicular, AD,  upon  the 
hypothenuse  BC,  what  re- 
lation do  the  two  triangles, 

ABD  and  ACD,  into  which  the  whole  triangle  is  di- 
vided, bear  to  each  other,  and  to  the  whole  triangle  ABC 
itself  ? 

A.    The  two  triangles,  ABD  and  ACD,  are  similar  to 
each  other,  and  to  the  whole  triangle  ABC. 
Q.  How  can  you  prove  this  1 

A.  The  triangle  ABD  is  similar  to  the  whole  triangle 
ABC,  because  the  two  triangles  being  both  right-angled, 
and  having  the  angle  at  x  common,  have  two  angles  in 
one  triangle,  respectively,  equal  to  two  angles  in  the  other 
(page  73,  case  1st) ;  and  for  the  same  reason  is  the  tri- 
angle ACD  similar  to  the  whole  triangle  ABC  (both 
being  right-angled,  and  having  the  angle  y  common) ; 
and  as  each  of  the  two  triangles,  ABD,  ACD,  is  similar 
to  the  whole  triangle  ABC,  these  two  triangles  must  be 
similar  to  each  other.  (Truth  II.) 

Q.  IVJiat  important  inferences  can  you  draw  from  the 
principle  you  have  just  established ? 

A.  1st.  In  the  two  similar  triangles,  ABD  and  ACD, 
the  sides  which  are  opposite  to  the  equal  angles,  must  be 


GEOMETRY.  75 

in  proportion  (condition  4th  of  geometrical  similarity, 
page  70) ;  and  we  shall  therefore  have  the  proportion 

BD:  AD=AD  :  DC;* 
that  is,  the  perpendicular  AD  is  a  mean  proportional 
between  the  two  parts  into  which  it  divides  the  hypothec 
nuse.  (Theory  of  proportion,  page  66.) 
A2%.  From  the  two  similar  triangles,  ABC,  ABD,  we 
shall  have  the  proportion 

BD  :  AB  =  AB  :  BC ; 
that  is,  the  side  AB,  of  the  right-angled  triangle  ABC, 
is  a  mean  proportional  between  the  whole  hypothenuse  BC, 
and  the  part  BD,  cut  off  from  it  by  the  perpendicular 
AD.f 

Sdly.  The  two  similar  triangles,  ACD  and  ABC,  give 
the  proportion 

DC  :  AC  =  AC  :  BC  ; 
that  is,  the  other  side,  AC,  of  the  right-angled  triangle 
ACD,  is   also  a  mean  proportional   between   the   whole 
hypothenuse  and  the  other  part,  DC,  cut  off" from  it  by  the 
perpendicular  AD. 

Remark.     The  five  last  queries  comprise  one  of  the  most  im 
portant  parts  of  Geometry.     The  principles  contained  in  them  are 
applied  to  the  solution  of  almost  every  geometrical  problem.     The 
beginner  will  therefore  do  well  to  render  himself  perfectly  familiar 
with  them. 

*  The  first  ratio  is  formed  by  the  two  sides,  BD  and  AD,  of  the 
triangle  ADB,  of  which  BD  is  opposite  to  the  angle  z,  and  AD  to 
the  angle  x  ;  and  the  second  ratio  is  formed  by  the  two  correspond- 
ing sides,  AD,  DC,  of  the  triangle  ADC ;  because  the  sides  AD, 
DC,  are  opposite  to  the  angles  y  and  u,  which  are  respectively 
equal  to  z  and  x. 

t  The  part  BD  of  the  hypothenuse,  situated  between  the  ex- 
tremity B  of  the  side  AB,  and  the  foot  D  of  the  perpendicular  AD, 
is  sometimes  called  the  adjacent  segment  to  AB.  (Legendre's 
Geometry,  translated  by  Professor  Farrar.) 


76  GEOMETRY. 

RECAPITULATION   OF  THE   TRUTHS  CONTAINED  IN 
THE  SECOND  SECTION. 


Ques.  Can  you  now  repeat  the  different  principles  re- 
specting the  equality  and  similarity  of  triangles,  which 
you  have  learned  in  this  section  ? 

Ans.  1.  If,  in  two  triangles,  two  sides  of  the  one  are 
equal  to  two  sides  of  the  other,  each  to  each,  and  the 
angles  which  are  included  by  them  are  also  equal  to  one 
another,  the  two  triangles  are  equal  in  all  their  parts,  that 
is,  they  coincide  with  each  other  throughout. 

2.  In  equal  triangles,  that  is,  in  triangles  which  coin- 
cide with  each  other,  the  equal  sides  are  opposite  to  the 
equal  angles. 

3.  If  one  side  and  the  two  adjacent  angles  in  one  tri- 
angle are  equal  to  one  side  and  the  two  adjacent  angles 
in  another  triangle,  each  to  each,  the  two  triangles  are 
equal,  and  the  angles  opposite  to  the  equal  sides  are  also 
equal.* 

4.  The  two  angles  at  the  basis  of  an  isosceles  triangle 
are  equal  to  one  another. 

5.  If  the  three  sides  of  one  triangle  are  equal  to  the 
three  sides  of  another,  each  to  each,  the  two  triangles 
coincide  with  each  other  throughout ;  that  is,  their  angles 
are  also  equal,  each  to  each. 

6.  In  every  triangle,  the  greater  side  is  opposite  to  the 
greater  angle,  and  the  greatest  side  to  the  greatest  angle. 

7.  In  a  right-angled  triangle,  the  greatest  side  is  oppo- 
site to  the  right  angle. 

*  This  principle,  though  already  demonstrated  in  the  first 
section,  is  repeated  here,  in  order  to  complete  what  is  said  on  the 
equality  of  triangles. 


GEOMETRY.  77 

8.  When  a  triangle  contains  two  equal  angles,  it  also 
has  two  equal  sides,  and  the  triangle  is  isosceles. 

9.  If  the  three  angles  in  a  triangle  are  equal  to  eac*. 
other,  the  sides  are  also  equal,  and  the  triangle  is  equi- 
lateral. 

10.  Any  one  side  of  a  triangle  is  smaller  than  the  sura 
of  the  two  other  sides. 

11.  If  from  a  point  within  a  triangle,  two  lines  are 
drawn  to  the  two  extremities  of  one  of  the  sides,  the 
angle  mader  by  those  lines  is  always  greater  than  the 
angle  of  the  triangle  which  is  opposite  to  that  side ;  but 
the  sum  of  the  two  lines,  which  make  the  interior  angle, 
is  smaller  than  the  sum  of  the  two  sides  which  include 
the  angle  of  the  triangle. 

12.  If  from  a  point  without  a  straight  line,  a  perpen- 
dicular is  let  fall  upon  that  line,  and,  at  the  same  time, 
other  lines  are  drawn  obliquely  to  different  points  in  the 
same  straight  line,  the  perpendicular  is  shorter  than  any 
of  the  oblique  lines,  and  is  therefore  the  shortest  line  that 
can  be  drawn  from  that  point  to  the  straight  line. 

13.  The  distance  of  a  point  from  a  straight  line  is 
measured  by  the  length  of  the  perpendicular,  let  fall  from 
that  point  upon  the  straight  line. 

14.  Of  several  oblique  lines  drawn  from  a  point  with- 
out a  straight  line,  to  different  points  in  that  straight  line^ 
that  one  is  the  shortest,  which  is  nearest  the  perpendicu- 
lar, and  that  one  is  the  greatest,  which  is  farthest  from 
the  perpendicular. 

15.  If  a  perpendicular  is  drawn  to  a  straight  line,  then 
two  oblique  lines  drawn  from  two  points  in  the  straight 
line,  on  each  side  of  the  perpendicular,  and  at  equal  dis- 
tances from  it,  to  any  one  point  in  that  perpendicular,  are 
equal  to  one  another. 

16.  If  a  perpendicular  is  drawn  to  a  straight  line,  there 

7* 


78  GEOxMETRY. 

is  but  one  point  in  the  straight  line,  on  each  side  of  the 
perpendicular,  such,  that  a  straight  line  drawn  from  it  to 
a  given  point  in  that  perpendicular,  is  of  a  given  length. 
17.  If  a  perpendicular  is  drawn  to  a  straight  line,  there 
is  but  one  point  in  the  straight  line,  on  each  side  of  the 
perpendicular,  from  which  a  line  drawn  to  a  given  point 
in  that  perpendicular,  makes  with  the  straight  line  an 
angle  of  a  required  magnitude. 

i  18.  If  two  sides  and  the  angle  which  is  opposite  to  the 
greater  of  them,  in  one  triangle,  are  equal  to  two  sides 
and  the  angle  which  is  opposite  to  the  greater  of  them  in 
another  triangle,  each  to  each,  the  two  triangles  coincide 
with  each  other  in  all  their  parts ;  that  is,  they  are  equal 
to  each  other. 

19.  If  the  hypothenuse  and  one  side  of  a  right-angled 
triangle,  are  equal  to  the  hypothenuse  and  one  side  of 
another  right-angled  triangle,  each  to  each,  the  two  right- 
angled  triangles  are  equal. 

20.  If  in  two  triangles  two  sides  of  the  one  are  equal  to 
two  sides  of  the  other,  each  to  each,  but  the  angle  in- 
cluded by  the  two  sides  in  one  triangle,  is  greater  than 
the  angle  included  by  them  in  the  other,  the  side  opposite 
to  the  greater  angle  in  the  one  triangle,  is  greater  than  the 
side  opposite  to  the  smaller  angle  in  the  other  triangle. 

21.  Every  parallelogram  is,  by  a  diagonal,  divided  into 
two  equal  triangles. 

22.  The  opposite  sides  of  a  parallelogram  are  equal  to 
each  other. 

23.  The  opposite  angles  in  a  parallelogram  are  equal 
to  each  other. 

24.  By  one  angle  of  a  parallelogram  a\\four  angles  are 
determined. 

25.  A  quadrilateral,  in  which  the  opposite  sides  are 
respectively  equal,  is  a  parallelogram. 


GEOMETRY.  719 

26.  A  quadrilateral,  in  which  two  sides  are  equal  and 
parallel,  is  a  parallelogram. 

27.  If  from  one  of  the  vertices  of  a  rectilinear  figure, 
diagonals  are  drawn  to  all  the  other  vertices,  the  figure  is 
divided  into  as  many  triangles  as  it  has  sides  less  two. 

28.  The  sum  of  all  the  angles  in  a  rectilinear  figure,  is 
equal  to  as  many  times  two  right  angles  as  the  figure  has 
sides  less  two. 


RECAPITULATION     OF    THE    TRUTHS    CONTAINED    IN 
PART  II. 

1.    On  Proportions. 

Ques.  1.  How  is  a  geometrical  ratio  determined  ? 

■Q.  2.  What  is  the  ratio  of  a  line  3  inches  in  length, 
to  a  line  of  12  inches?  What,  the  ratio  of  a  line  2  inches 
in  length,  to  one  of  10  inches,  &c? 

Q.  3.  When  two  geometrical  ratios  are  equal  to  one 
another,  what  do  they  form  ? 

Q.  4.  What  is  a  geometrical  proportion  ? 

Q.  5.  What  signs  are  used  to  express  a  geometrical 
proportion  ? 

Q.  6.  What  sign  is  put  between  the  two  terms  of  a 
ratio  1 

Q.  7.  What  sign  is  put  between  the  two  ratios  of  a 
proportion  ? 

Q.  8.  What  are  the  first  and  fourth  terms  of  a  geomet- 
rical proportion  called  ? 

Q.  9.  What  are  the  second  and  third  terms  of  a  geo- 
metrical proportion  called  1 

Q.  10.  What  are  the  most  remarkable  properties  of 
geometrical  proportions  ? 


80  GEOMETRY. 

Ans.  a.  In  every  geometrical  proportion  the  two  ratios 
may  be  inverted, 

b.  In  every  geometrical  proportion  the  order  of  the 
means  or  extremes  may  be  inverted. 

c.  If  two  geometrical  proportions  have  a  ratio  common, 
the  two  remaining  ratios  make  again  a  proportion. 

d.  If  you  have  several  geometrical  proportions,  of 
which  the  second  has  a  ratio  common  with  the  first,  the 
third  a  ratio  common  with  the  second,  the  fourth  a  ratio 
common  with  the  third,  &,c.,the  sum  of  all  the  first  terms 
will  be  in  the  same  ratio  to  the  sum  of  all  the  second 
terms,  as  the  sum  of  all  the  third  terms  is  to  the  sum 
of  all  the  fourth  terms ;  that  is,  the  sums  make  again  a 
proportion. 

e.  The  second  term  of  a  proportion  being  added  once, 
or  any  number  of  times,  to  the  first  term,  and  the  fourth 
term  the  same  number  of  times  to  the  third  term,  they 
will  still  be  in  proportion  ;  and  in  the  same  manner  can 
the  first  term  be  added  a  number  of  times  to  the  secoud 
term,  and  the  third  the  same  number  of  times  to  the 
fourth  term,  without  destroying  the  proportion. 

f.  The  second  term  may  also  be  once,  or  any  number 
of  times,  subtracted  from  the  first- term,  and  the  fourth 
from  the  third  term,  without  destroying  the  proportion  ; 
or  the  first  term  may  also  be  subtracted  from  the  second, 
and  the  third  from  the  fourth — and  the  result  will  still  be 
a  geometrical  proportion. 

g.  If  all  the  terms  of  a  geometrical  proportion  are 
multiplied  or  divided  by  the  same  number,  the  proportion 
remains  unaltered. 

h.  From  three  terms  of  a  geometrical  proportion  the 
fourth  term  can  be  found. 

t.  If  four  lines  are  together  in  a  geometrical  propor- 


GEOMETRY.  gl 

tion,  their  lengths,  expressed  in  numbers  of  rods,  feet, 
inches,  &.C.,  are  in  the  same  proportion. 

k.  In  every  geometrical  proportion,  the  product  of  the 
two  mean  terms  is  equal  to  that  of  the  two  extremes. 

I.  When  the  two  mean  terms  of  a  geometrical  proport 
tion  are  equal  to  each  other,  either  of  them  is  called  a 
mean  proportional  between  the  two  extremes. 


QUESTIONS  ON  SIMILARITY  OF  TRIANGLES. 

Ques.  What  other  principles  do  you  recollect  in  the 
second  part  of  the  second  section  1 

Ans.  1.  If  one  side  of  a  triangle  is  divided  into  any 
number  of  equal  parts,  and  then,  from  the  points  of 
division,  lines  are  drawn  parallel  to  one  of  the  two  other 
sides,  the  side  opposite  to  the  one  that  has  been  divided 
will,  by  these  parallels,  be  divided  into  as  many  equal 
parts  as  the  first  side. 

2.  If,  in  a  triangle,  a  line  is  drawn  parallel  to  one  of 
the  sides,  that  parallel  divides  the  two  other  sides  into 
such  parts  as  are  in  proportion  to  each  other  and  to  the 
whole  of  the  two  sides  themselves ;  and  the  reverse  of  this 
principle  is  also  tf  ue  ;  namely,  a  line  must  be  parallel  to 
one  of  the  sides  of  a  triangle,  if  it  divides  the  two  other 
sides  proportionally. 

3.  If,  in  a  triangle,  a  line  is  drawn  parallel  to  one  of 
the  sides,  the  triangle  which  is  cut  off  by  it  is  similar  to 
the  whole  triangle. 

4.  If  the  three  angles  in  one  triangle  are  equal  to  the 
three  angles  in  another  triangle,  each  to  each,  the  two 
triangles  are  similar  to  one  another ;  and  the  same  is  the 
case  if  only  two  angles  in  one  triangle  are  equal  to  two 
angles  in  another,  each  to  each. 


82  GEOMETRY. 

5.  If  an  angle  in  one  triangle  is  equal  to  an  angle  in 
another,  and  the  two  sides  which  include  that  angle  in 
the  one  triangle  are  in  proportion  to  the  two  sides  which 
include  the  equal  angle  in  the  other,  these  two  triangles 
are  similar  to  each  other. 

6.  If  the  three  sides  of  one  triangle  are  in  proportion 
'o  the  three  sides  of  another,  the  two  triangles  are  similar 
to  each  other.* 

7.  If,  in  a  right-angled  triangle,  a  perpendicular  is  let 
fall  from  the  vertex  of  the  right  angle  upon  the  hypothe- 
nuse,  that  perpendicular  divides  the  whole  of  the  triangle 
into  two  parts,  which  are  similar  to  the  whole  triangle, 
and  to  each  other. 

8.  The  perpendicular  let  fall  from  the  vertex  of  a  right- 
angled  triangle  upon  the  hypothenuse,  is  a  mean  pro- 
portional between  the  parts  into  which  it  divides  the 
hypothenuse. 

9.  In  every  right-angled  triangle,  each  of  the  sides 
which  include  the  right  angle  is  a  mean  proportional 
between  the  hypothenuse  and  that  part  of  it,  which  lies 
between  the  extremity  of  that  side  and  the  foot  of  the 
perpendicular  let  fall  from  the  vertex  of  the  right  angle 
upon  the  hypothenuse. 


*  The  teacher  will  do  well  to  let  the  pupil  repeat  the  different 
eases  where  two  triangles  are  similar  to  each  other.  (Page   73.) 


SECTION  III. 


OF  THE  MEASUREMENT  OF  SURFACES. 


Preliminary  Remarks.  We  determine  the  length  of  a  line, 
by  finding  how  many  times  another  line,  which  we  take  for  the 
measure,  is  contained  in  it.  The  line  which  we  take  for  the 
measure  is  chosen  at  pleasure  ;  it  may  be  an  inch,  a  foot,  a  fathom, 
a  mile,  &c.  If  we  have  a  line  upon  which  we  can  take  the  length 
of  an  inch  3  times,  we  say  that  line  measures  3  inches,  or  is  3 
inches  long.  In  like  manner,  if  we  have  a  line  upon  which  we 
can  take  the  length  of  a  fathom  3  times,  we  call  that  line  3  fath- 
oms, &c.  To  find  out  which  of  two  lines  is  the  greater,  we  must 
measure  them.  If  we  take  an  inch  for  our  measure,  that  line  is 
the  greater,  which  contains  the  greater  number  of  inches.  If  we 
take  a  foot  for  our  measure,  that  line  is  the  greater,  which  contains 
the  greater  number  of  feet,  &c. 

To  measure  the  extension  of  a  surface,  we  make  use  of  another 
surface,  commonly  a  square  (  □  ),  and  see  how  many  times  it  can 
be  applied  to  it ;  or,  in  other  words,  how  many  of  those  squares  it 
takes  to  cover  the  whole  surface.  The  length  of  the  square  side  is 
arbitrary.  If  it  is  an  inch,  the  square  of  it  is  called  a  square  inch  ; 
if  it  is  a  foot,  a  square  foot ;  if  it  is  a  mile,  a  square  mile,  &c.  The 
extension  of  a  surface,  expressed  in  numbers  of  square  miles,  rods 
feet,  inches,  &c,  is  called  its  area. 

Rcinark  2.  If  we  take  one  of  the 
sides  of  a  triangle  for  the  basis,  the 
perpendicular  let  fall  from  the  vertex 
of  the  opposite  angle,  upon  that  side, 
is  called  the  altitude  or  height  of  the 
triangle. 

If,  in  the  triangle  ABC,  (Fig.  I.) 
for  instance,  we  call  AC  the  basis,  the 
perpendicular  BD  will  be  its  height.  If 
the  perpendicular  BD  should  fall  with- 
out the  triangle  ABC  (as  in  Fig.  II.), 


Fig.  I. 


84  GEOMETRY. 

we  need  oiily  extend  the  basis,  and  then  let  fall  the  perpendicular 
BD  upon  its  farther  extension  CD. 

*  M a * & 


a* 


2sT  V      0 


$ 


If,  in  a  parallelogram,  ABCD,  we  take  AD  for  the  basis,  any  per 
pendicular,  MN,  CO,  PQ,  &c.,  let  fall  from  the  opposite  side  BC, 
or  its  farther  extension  CR,  upon  tbat  basis,  or  its  farther  extension 
DS,  measures  the  height  of  the  parallelogram.  For  in  a  parallelo- 
gram the  opposite  sides  are  parallel  to  each  other  (see  Definitions), 
and  all  the  perpendiculars,  let  fall  from  one  of  two  parallel  lines  to 
the  other,  are  equal  (Query  11,  Sect.  I.).  What  in  this  respect  holds 
of  a  parallelogram  is  applied  also  to  a  square,  a  rhombus,  and  a 
rectangle^;  for  these  three  figures  are  only  modifications  of  a  paral- 
lelogram.    (See  Definitions.) 

As  in  every  rectangle,  ABCD,  the  adjacent      (*. T) 

sides,  AB,  BD,  are  perpendicular  to   each I  _ 

other,  it  is  evident  that  if  AB  is  taken  for  the 

basis,  the  side  BD  itself  is  the  height  of  the  rectangle. 

Remark  3.  We  call  two  geometrical  figures  equal*  to  one 
another,  when  they  have  equal  areas  (see  preliminary  remark  to 
Sect.  II.).  Thus  a  triangle  is  said  to  be  equal  to  a  rectangle  when 
it  contains  the  same  number  of  square  miles,  rods,  feet,  inches,  &c, 
as  that  rectangle. 


*  The  term  equivalent  would  undoubtedly  be  better  ;    but  as 
there  is  no  generally  adopted  sign  in  mathematics  to  express  that 
two  things  are  equivalent  without  being  exactly  the  same,  we  are 
~°  the  term  equal. 


GEOMETRY. 


85 


D  1    2 


4    5 


L 


6  C 
4 


12    3    4     5    6 


QUERY  I 

If  the  basis,  AJB,  of  a  rectan- 
gle, ABCD,  measures  6  inches, 
and  the  height,  the  side  BC,  4 
inches,  how  many  square  inches 
are  there  in  the  rectangle  ? 

A.    Twenty-four. 

Q.  How  can  you  prove  this  ? 

A.  If  a  rectangle  is  four  inches  high,  I  can  divide  it, 
like  the  rectangle  ABCD  (see  the  figure),  into  four  rect- 
angles, each  of  which  ,is  one  inch  high,  and  has  its  basis 
equal  to  the  basis  of  the  whole  rectangle.  And  as  the 
basis,  AB,  of  the  rectangle  measures  6  inches,  by  raising 
upon  it,  at  the  distance  of  an  inch  from  each  other,  the 
perpendiculars  1,  2,  3,  4,  5,  each  of  the  four  rectangles 
will  be  divided  into  6  square  inches  ;  and  therefore  the 
whole  rectangle  ABCD  into  24  square  inches. 

Q.  How  many  square  inches  are 
there  in  a  rectangle,  whose  basis  is 
5,  and  height  3  inches  ? 

A.  Fifteen.  Because  in  this  case 
I  can  divide  the  rectangle  into  3 
rectangles  of  5  square  inches  each. 

Q.  Supposing  the  measurements  of  the  first  rectangle 
were  given  in  feet,  in  rods,  or  in  miles,  instead  of  inches, 
how  many  square  feet,  rods,  or  miles  would  there  be  in 
the  rectangle  1 

A.  If  its  measurements  were  given  in  feet,  it  would 
contain  24  square  feet ;  if  they  were  given  in  rods,  it 
would  contain  24  square  rods,  &c.  ;  for  in  these  cases  I 
need  only  imagine  the  lines,  1,  2,  3,  4,  &c,  to  be  drawn 
a  foot,  a  rod,  &c,  apart ;  the  number  of  divisions  will 
8  "  . 


C     1     2     3     4     5JD 


3 
2 

1 


12      3     4     5 


86 


GEOMETRY. 


remain  the  same ;  nothing  but  their  size  will  be  altered. 
And  the  same  reasoning  applies  to  the  second  rectangle.* 

Q.  Can  you  now  give  a  general  rule  for  finding  the 
area  of  a  rectangle  ? 

A.  Yes.  Multiply  the  length  of  the  basis  given  in  rods, 
feet,  inches,  fyc,  by  the  height  expressed  in  units  of  the 
same  kind. 

Q.  Can  you  now  tell  me  how  to  find  the  area  of  a 
square  ? 

A.  The  area  of  a  square  is  found  by  multiplying  one 
of  its  sides  by  itself  For  a  square  is  a  rectangle  whose 
sides  are  all  equal  (see  Definitions)  ;  and  the  area  of  a 
rectangle  is  found  by  multiplying  the  basis  by  an  adja- 
cent side. 


QUERY  II. 

If  a  parallelogram  ABEF 
stands  on  the  same  basis,  AB, 
as  a  rectangle,  ABCD,  and 
has  its  height  equal  to  the 
height  of  that  rectangle,  what 
relation  do  the  areas  of  these 
two  figures  bear  to  each  other  ? 

A.  The  area  of  the  parallelogram  ABEF  is  equal  to 
the  area  of  the  rectangle  ABCD ;  therefore  I  can  say 
that  the  parallelogram  ABEF  is  equal  to  the  rectangle 
ABCD  (see  remark  3d,  Introd.  to  Sect.  III.). 


C         D     F 

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*  The  teacher  may  also  give  his  pupils  a  rect- 
angle whose  measurements  are  both-  given  in 
fractions  ;  for  instanc*era  rectangle  of  3£  inches 
in  length  and  2^  inches  high,  and  then  show  by 
the  figure  that  this  rectangle  measures  6  square 
inches,  2  half  square  inches,  |  and  J  of  a  square  inch;  in  the  whole 
7}  square  inches,  which  is  the  answer  to  the  multiplication  of  3J 


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GEOMETRY.  87 

Q.  How  can  you  prove  it  1 

A.  The  right-angled  triangle  ACF  has  the  hypothe- 
nuse  AF  and  the  side  AC,  equal  to  the  hypothenuse  BE 
and  the  side  BD,  in  the  right-angled  triangle  BDE,  each 
to  each  (AF  and  BE,  AC  and  DB,  being  opposite  sides 
of  the  parallelogram  ABEF,  and  the  rectangle  ABCD, 
respectively)  ;  therefore  these  two  triangles  are  equal 
(page  47)  i  and  by  taking  from  each  of  the  two  equal  tri- 
angles ACF,  BDE,  the  part  DGF  common  to  both,  the 
remainders,  AGDC,  BGFE,  are  also  equal(Truth  IV.); 
and  then,  by  adding  again  to  each  of  the  equal  remainders 
the  same  triangle  ABG,  the  sums,  that  is,  the  rectangle 
ABCD  and  the  parallelogram  ABEF  are  equal  to  one 
another.    (Truth  III.) 

Q.  What  important  truths  can  you  infer  from  the  one 
you  have  just  demonstrated  ? 

A.  1st.  All  parallelograms,  which  have  equal  bases  and 
heights,  are  equal  to  one  another ;  for  each  of  them  is 
equal  to  a  rectangle  upon  the  same  basis,  and  of  the  same 
lieight.  (Truth  I.) 

2dly.  Parallelograms  upon  equal  bases,  and  between  the 
same  parallels,  am  equal  to  one  another  ;  for  if  they  are 
between  the  same  parallels,  their  heights  must  be  equal, 
(Query  11,  Sect.  I.) 

3dly.  The  area  of  a  parallelogram  is  found  by  multi- 
plying the  basis,  given  in  rods,,  feet,  inches,  fyc,  by  the 
Jteight,  expressed  in  units  of  the-same  kind.  Because  the 
area  of  the  rectangle  upon  the  same  bas-is  and  of  the  same 
height  to  which  it  is  equal,  is  found  in  the  same  manner. 

4thly.  The  area  of  a  rhombus  or  lozenge  is  found  like 
that  of  a  parallelogram,  a  lozenge  being  only  a  peculiar 
hind  of  parallelogram. 

5thly.  The  areas  of  parallelograms  are  to  each  other, 
as  the  products  obtained  by  multiplying  the  length  of  the 


m 


88  GEOMETRY. 

bases  of  the  parallelograms  by  their  heights;  because 
these  products  are  the  areas  of  the  parallelograms. 

The    parallelogram 
ABCD,  for   instance,  *pf- — ty   E_0__F 

is  to  the  parallelogram  /  /      /  ~7 

GHEF,  as  the  product         Z L/     /_ !__/ 

of  the  basis  AB,  by  the     A         NB    a  rH 

height  MN,  is  to  the  product  of  the  basis  GH,  by  the 
height  OP  ;  because  AB  multiplied  by  MN  is  the  area  of 
the  parallelogram  ABCD,  and  GH  multiplied  by  OP  is 
the  area  of  the  parallelogram  GHEF.  This  proportion 
may  be  expressed  thus  : 

Parallelogram  ABCD  :  parallelogram  GHEF  =  AB 
X  MN  :  GH  x  OP. 

6thly.  Rectangles  or  parol-  . 

hlograms,   which    have    equal  / — j j    „    1 

bases,  are  to  each  other  as  their          /           /     /     )    ~J 
heights.  L U      L \J 

For  if,  in  the  above  propor-     A      NB  G     Pir 
tion,  the  basis  AB  is  equal  to  the  basis  GH,  you  can  write 
AB  instead  of  GH,  and  thereby  change  it  into 

Parallelograms  ABCD  :  parallelograms  GHEF  = 

AB  X  MN  :  AB  X  OP  ; 

that  is,  the  parallelogram  ABCD  is  to  the  parallelogram 

GHEF,  as  AB  times  the  height  MN  is  to  AB  times  the 

height  OP  ;   or,  which  is  the  same,  as  the   height  MN 

alone  is  to  the  height  OP  alone  ;  which  is  written  thus: 

Parallelogram  ABCD  :  parallelogram  GHEF  = 

MN  :  OP. 

7thly.  In  precisely  the  same  manner  it  may  be  proved, 
that  if  the  heights  MN  and  OP  are  equal,  the  parallelo- 
grams ABCD,  GHEF,  are  to  each  other  as  their  bases; 
which  may  be  expressed  thus : 

Parallelogram  ABCD  :  parallelogram  GHEF  = 
AB  :  GH. 


GEOMETRY.  89 

QUERY  III. 

If  two  triangles,  ABC,  ABE,  stand  on  the  same  basis 
AB,  and  have  equal  heights  CK,  EG,  what  relation  do 
the  areas  of  these  triangles  bear  to  each  other  ? 

nam  j2r 


A.   The  areas  of  these  triangles  are  equal. 

Q.  How  can  you  prove  it  1 

A.  Draw  the  line  AD  parallel  to  BC ;  BF  parallel  to 
AE  ;  and  through  the  two  vertices  G  and  E,  the  line  DF 
parallel  to  AG  (which  is  possible  since  the  heights  CK 
and  EG  are  equal).  The  area  of  the  parallelogram 
ABCD  will  be  equal  to  the  area  of  the  parallelogram 
ABEF  (Query  2,  Sect.  III.)  ;  and  as  the  triangle  ABC 
is  half  of  the  parallelogram  ABCD  (Query  12,  Sect.  II.), 
and  the  triangle  ABE  half  of  the  parallelogram  ABEF, 
the  areas  of  these  two  triangles  are  also  equal  to  one 
another  ;  for  if  the  wholes  are  equal,  the  halves  must  be 
equal ;  and  in  the  same  way  it  may  be  proved  that  trian- 
gles, which  have  equal  bases  and  heights,  are  equal  to 
one  another. 

Q.  What  consequences  follow  from  the  principle  just 
advanced  1 

A.  1st.  Every  triangle  is  half  of  a  parallelogram  upon 
equal  basis  and  of  the  same  height.  (This  is  evident  from 
looking  at  the  figure,  and  from  Query  12,  Sect.  II.) 

2d.  The  area  of  a  triangle  is  half  of  the  area  of  a 
parallelogram  upon  the  same  basis  and  of  the  same  height. 
Thus  the  area  of  a  triangle  is  found  by  multiplying  its 
■     8* 


90  GEOMETRY. 

basis  by  its  height,  and  dividing  the  product  by  2  ;*  for 
the  area  of  a  parallelogram  is  equal  to  the  whole  pro-duct 
of  the  basis  by  the  height.f 

3d.  The  areas  of  triangles  upon  the  same  basis  and 
between  the  same  parallels  are  equal;  because  if  they  are 
between  the  same  parallels,  their  heights  are  equal ;  and 
we  have  the  same  case  as  in  the  last  query  ;  namely,  tri- 
angles upon  the  same  basis,  and  of  equal  heights. 

4th.  The  areas  of  triangles  are  to  each  other  as  the 
products  of  their  bases  by  their  heights  :  for  the  halves  of 
these  products  being  the  areas  of  the  triangles,  the  whole 
products  must  be  in  the  same  ratio.  Thus  the  area  of 
the  triangle  ABC  is 
to  the  area  of  the 
triangle  EGP,  as  the 

basis  AB  multiplied     Ji^- jy-  \B  JE7 

by  the  height  CN,  is 

to  the  basis  EG,  multiplied  by  the  height  PM  ;  which 

may  be  expressed  thus  : 

Triangle  ABC  :  triangle  EGP  =  ABx  CN  :  EG  X  PM. 

5th.  Hie  areas  of  triangles  upon  equal  bases  are  to 
each  other  as  the  heights  of  the  triangles  ;  because  the 
areas  of  parallelograms  upon  the  same  bases  and  of  the 
same  heights,  are  to  each  other  in  the  ratio  of  the  heights ; 
and  their  halves  (the  areas  of  the  triangles)  must  be  in 


*  Instead  of  multiplying  the  basis  by  the  whole  height  and 
dividing  the  product  by  2,  you  may  multiply  the  basis  by  half  the 
height,  or  the  height  by  half  the  basis. 

t  If  the  basis  of  a  triangle  is  8  feet  and  the  height  4  feet,  the 
area  of  the  triangle  is  equal  to  4  times  8,  divided  by  2;  that  is, 
16  square  feet ;  whereas  the  rectangle  upon  8  feet  basis  and  4 
feet  high,  measures  32  square  feet,  which  is  double  the  area  of  the 
triangle. 

1 


GEOMETRY.  01 

the  same  ratio.*     Thus  if  the  two  triangles  ABG,  ECF, 
O 
/i\  F 


have  their  bases  AB,  EC,  equal  to  each  other,  we  have 
the  proportion : 

Triangle  ABG  :  triangle  ECF  —  CM  :  FN. 
6th.  The  areas  of  triangles,  which  have  equal  heights, 
are  to  each  other  as  the  bases  of  the  triangles.  This  truth 
follows  like  the  preceding-,  one  from  the  same  principle 
established  with  regard  to  parallelograms,  of  which  the 
triangles  are  the  halves.  (Page  88,  7thl.y.) 

QUERY  IV. 
How  do  you  find  the  -A J?    j£ 


3 


area  of  a  trapezoid  ? 

A.    By    multiplying     . 
the  sum  of  the  two  par-  ~  I?  D 

allel  sides  by  their  distance,  and  dividing  the  product 
by  2. 

Q.  How  can  you  prove  this  1 

A.  By  drawing  the  diagonal  AD,  the  trapezoid  ABCD, 
will  be  divided  into  the  two  triangles  ACD  and  ABD. 
The  area  o'fthe  triangle  ACD  is  found  by  multiplying  its 
basis,  CD,  by  its  height  AF,  and  dividing  the  product 
by  2.  (Page  89,  2d.)  In  the  same  manner  we  find  the 
area  of  the  triangle  ABD,  by  multiplying  its  basis,  AB, 

*  This  principle  and  the  following  one  might  have  been  estab- 
lished immediately  from  the  proportion  : 

Triangle  ABC  :  triangle  EGP  =  AB  X  CN  :  EG  X  PM, 
in  precisely  the  same  manner,  as  it  has  been  proved  for  parallelo- 
grams.   (Page  88, 6thly.) 


92  GEOMETRY. 

by  its  height  DE,  and  dividing  the  product  by  2;  and 
as  the  height,  DE,  of  the  triangle  ABD,  is  equal  to 
the  height  AF,  of  the  triangle  ACD  (because  DE  and 
AF  are  perpendiculars  between  the  same  parallels),  we 
can  find  the  area  of  the  two  triangles,  or  of  the  whole 
trapezoid,  ABCD,  at  once,  by  multiplying  the  sum  of  the 
two  parallel  lines  AB,  CD,  by  their  distance  AF,  and 
dividing  the  product  by  2.* 

QUERY  V. 

Hoic  do  you  find  the  area  of,- 
a  polygon  ABCDEF,  or-,  in 
general,  of  any  other  rectilinear 
figure  ? 

A.  By  dividing  it  by  means 
of  -diagonals   [as  in  the  figure  ~~    •& 

before  you),  or  by  any  other  means  into  triangles.  The 
area  of  each  of  these  triangles  is  then  easily  found  by  the 
rule  given  (page  89,  2d.)  ;  and  the  sum  of  the  areas  of 
all  the  triangles,  into  which  the  figure  is  divided,  is  the 
area  of  it.  . 


*  If  you  multiply  two  numbers  successively  by  the  same  number, 
and  then  add  the  products  together,  the  answer  will  be  the  same 
as  the  sum  of  the  two  numbers  at  once  multiplied  by  that  number. 
Multiply  each  of  the  numbers,  6  and  5,  for  instance,  by  4,  and 
then  add  the  products,  24  and  20,  together,  you  will  have  44  ;  and 
adding,  in  the  first  place,  6  to  5,  and  then  multiplying  the  sum, 
11,  by  4,  you  will  again  have  44. 

Instead  of  multiplying  the  sum  of  the  two  parallel  sides  by  their 
distance,  and  then  dividing  the  product  by  2,  you  may  multiply,  at 
once,  half  the  sum  of  the  two  parallel  sides  by  their  distance  ;  or 
the  sum  of  the  two  parallel  sides  by  half  their  distance. 


GEOMETRY. 


93 


UUERY  VI. 
If,  upon  each  of  the  three  sides  AB,  AC,  BC,  of  a 
right-angled  triangle  ABC,  you  construct  a  square ,  what 
relation  do  the  squares  constructed  upon  the  sides  ACy 
BC,  bear  to  the  square  constructed  upon  the  hypothenusey 
AB1 


A.  The  square  ABHK,  constructed  upon  the  hypothe- 
nuse  AB,  equals,  in  area,  the  two  squares  ACDE, 
BCGF,  constructed  upon  the  two  sides  AC,  BC. 

Q.  How  can  you  prove  it  by  this  diagram,  in  which 
the  perpendicular  CM,  is  let  fall  from  the  vertex  C,  of 
the  right-angled  triangle  ABC,  upon  the  hypothenuse 
AB,  and  extended  until,  in  I,  it  meets  the  side  HK, 
opposite  to  the  hypothenuse ;  and  DB  and  CH,  are 
joined?        v  :  . 

A.  In  the  first  place,  I  should  remark  that  the  two 
sides  AB,  AD,  of  the  triangle  ABD,  are  equal  to  the  two 
sides  AH,  AC,  of  the  triangle  ACH,  each  to  each  (AH 


94  GEOMETRY. 

and  AB,  being  sides  of  the  same  square,  ABHK ;  and, 
AC  and  AD,  being  sides  of  the  square  ACDE) ;  and 
that  the  angle  DAB,  included  by  the  sides  AD,  AB,  is 
also  equal  to  the  angle  CAH,  included  by  the  two  sides 
AC,  AH  (for  each  of  these  angles  is  formed  by  the  angle 
CAB  being  added  to  the  right  angle  of  a  square)  ;  there- 
fore these  two  triangles  are  equal  to  each  other,  (Query  1, 
Sect.  II.) 

•  Q.  Having  proved  that  the  triangle  ABD  is  equal  to 
the  triangle  ACH,  what  can  you  infer  from  it? 

A.  That  the  area  of  the  square  ACDE,  is  equal  to  the 
area  of  the  rectangle  AHIM.  For  the  area  of  the  triangle 
ABD,  is  half  of  the  area  of  the  square  ACDE ;  because 
the  triangle  ABD  stands  upon  the  same  basis  AD,  as  the 
square  ACDE,  and  has  its  height  BQ,,  equal  to  the  height 
AC  of  that  square ;  and  it  has  been  proved  that  the"  area 
of  every  triangle  is  half  of  the  area  of  a  rectangle  or 
square  of  equal  basis  and  height.  (Page  89,  1st.)  For  the 
same  reason  is  the  area  of  the  triangle  ACH,  equal  to 
half  the  area  of  the  rectangle  AHIM  ;  for  the*  triangle 
ACH,  stands  on  the  same  basis  AH,  as  the  rectangle 
AHIM,  and  has  its  height  CO.  equal  to  the  height  AM, 
of  that  rectangle ;  and  as  the  halves ,  the  two  triangles 
ABD  and  ACH,  are  eqnal  to  eacrf  other,  the  wholes,  the 
squares  ADEC  and  AHIM,  must  also  be  equal  to  each 
other,  fn  precisely  the  same,  manner  I  can  prove 
from  the  equality  of  the  two  triangles  ABG  and  BCK, 
that  the  square  BCFG  is^equal  to  the  rectangle  'MBIK  ; 
and  because  the  area  of  the  square  ADEC,  is  equal  to 
the  area  of  the  rectangle  AHIM,  and  the  area  of  the 
square  BCFG  is  equal  to  the  area  of  the  rectangle 
MBIK  ;  therefore  the  sum  of  the  areas  of  the  two  rectan- 
gles, AHIM  and  MBIK,  that  is,  the  area  of  the  square 
upon  the  hypothcnuse  AB,  is  equal  to  the  sum  of  the 


GEOMETRY.  95 

areas  of  the  squares  constructed  upon  the  two  sides  AC, 
BC. 


Remark.  For  the  discovery  of  this  principle,  we  are  indebted  to 
Pythagoras,  a  famous  Greek  mathematician.  It  is  a  very  important 
one,  and  teaches  how  to  find  one  of  the  sides  of  a  right-angled  tri- 
angle when  the  two  others  are  given*  if,  for  instance,  the  two 
sides  AC,  BC,  of  the  right-angled  triangle  ABC,  were  known  to 
measure,. one  5,  the  other  6  inches,  the  sum  of  their  squares  25 
(5  time3^5.),.and  36  (6  times  6),  equal  to  61,  would  be  the  area  of 
the  square  of  the  hypothenuse  ;  and  the  square  root  of  that  number 
would  be  the  hypothenuse  AB,  itself.  If  the  hypothenuse  and  one 
of  the  sides  are  given,  -you  need  only  subtract  the  square  of  the 
side  from  the  square  of  the  hypothenuse,  and  then  the  square  root 
of  the  remainder  is.  the  other  side.  If,  for  instance,  the  hypothe- 
nuse of  a  right-angled  triangle  were  10  feet,  and  one  of  the  sides 
6  feet ;  the  square  of  the  hypothenuse  would  be  10  times  10,  or  100, 
and  the  square  of  6,  which  is  36,  subtracted  from  100,  ldaves*64, 
which  would  be  the. square  of  the  side  to  be  found  ;  and  taking  the 
square  root  of  it,  which  is  8  (because  8  times  "8  are  64),  you  will 
have  the  side  itself* 

QUERY  VII. 

It  has '  been  proved  (page  90,  4th),  that  the  areas  of 
any  two  triangles  are  to  each  other  as  their  bases,  multi- 
plied by  their  heights  ;  can  you  now  find  out  the  relation 
which  the  bases  and  heights  of  similar  triangles  bear  to 
each  other  ? 

A.  In  similar  triangles  the  bases  are  in  proportion  to 
the  heights. 

Q.  How  can  you  prove  this  ? 


*  We  shall  hereafter  give   the  geometrical  solutions  of  these 
problems. 


9(5 


GEOMETRY. 


A.  Let  there  be  any  two  similar  triangles  ABC,  AED. 
Place  the  smaller  one  ABC,  upon  the  larger  AED,  in 


such  a  way  that  the  angle  at  A  falls  upon  the  angle  at  A, 
and  from  the  vertices,  C  and  D,  let  fall  the  perpendicu- 
lars CM,"  DO,  upon  AO.  Then  the  two  triangles  BCM, 
EDO,  are  both  right-angled,  and  the  angle  CBM  is  equal 
to  the  angle  DEO  (because  in  the  two  similar  triangles 
ABC,  AED,  the  angles  ABC  and  AED,  are  equal  to 
each  other,  and  CBM  and  DEO,  make  with  them,  respec- 
tively, two  right  angles)  ;  therefore  the  third  angle  BCM 
in  the  triangle  BCM,  is  also  equal  to  the  third  angle 
EDO,  in  the  triangle  EDO,  and  the  two  triangles  BCM, 
EDO,  are  similar.  (Page  73,  1st.)  But  in  similar  trian- 
gles the  sides  opposite  to  the  equal  angles  are  propor- 
tional, consequently  we  have 

CM  :DO  =  CB  :  DE; 
and  in -the  similar- triangles  ABC,  ADE,     • 
AB  :  AE^CB  :  DE. 

These  two  proportions  have  the  second  ratio  common ; 
therefore  the  two  first  ratios  must  again  make  a  propor- 
tion (Theory  of  Proportions,  Principle  3d.),  namely: 
AB  :  AE  =  CM  :  DO. 

This  proportion  expresses  that  the  bases  AB  of  the 
smaller  triangle  ABC,  is  to  the  bases  AE  of  the  larger 
triangle  AED,  as  the  height  CM,  of  the  first  triangle 
ABC,  is  to  the  height  DO,  of  the  triangle  AED. 


GEOMETRY. 


97 


QUERY  VIII. 

From  what  you  have  learned  in  the  preceding  query, 
can  you  determine  the  proportion  zohich  the  areas  qfsimi* 
lar  triangles  bear  to  each  other  ? 

A.  The  areas  of  similar  triangles  are  to  each  other  as 
the  squares  upon  the  corresponding  sides. 


li 


r 

a^ 

^^  , 

BM    E  O 


Q.  How  can  you  prove  this,  for  instance,  of  the  two 
similar  triangles  ABC,  ABD  ? 

A.  Let  us  place  the  smaller  triangle  ABC  upon  the 
larger  AED,  as  in  the  last  query;  and  upon  AB  and 
AE,  construct  the  squares  ABST,  AERP.  Then  the 
triangles  ABC,  AED,  have  the  same  bases  AB,  AE,  as 
the  triangles  ABT,  AEP,  and  their  heights  CM,  DO,  are 
in  proportion  to  the  heights  TB,  PE,  of  the  triangles 
ABT,  AEP  (TB  and  PE  being  respectively  equal  to 
AB,  AE,  which,  in  the  last  query,  are  proved  to  be  pro- 
portional to  CM  and  DO) ;  therefore  the  areas  of  the 
triangles  ABC,  AED,  are  in  proportion  to  the  areas  of 
the  triangles  ABT,  AEP.  (Page  90,  5thly.)  But  if  the 
two  triangles  ABC,  AED,  are  in  proportion  to  the  two 
triangles  ABT,  AEP,  which  are  the  halves  of  squares 
ABST,  AERP,  they  must  also  be  in  proportion  to  the 
squares  themselves  ;  which  may  be  expressed  thus : 
9 


I 


98 


GEOMETRY. 


Triangle  ABC  :  triangle  AED  =  AB  X  AB  :  AE  X  AE, 
and  is  read  : 

The  area  of  the  triangle  AED  is  as  many  times  greater 
than  the  area  of  the  triangle  ABC,  as  the  area  of  the 
square  upon  the  side  AE,  is  greater  than  the  area  of  the 
square  upon  the  corresponding  side  AB. 

Q.  Can  you  prove  that  the  same  ratio  exists  also  be- 
tween the  squares  upon  the  sides  AC  and  AD,  and  also 
between  the  two  sides  CB,  DE,  of  the  similar  triangles 
ABC,  AED  1 

A.  Yes.  For  to  prove  it  of  the  two  sides  AC  and 
AD,  I  need  only  take  them  for  the  bases  of  the  two  trian- 
gles ;  and  to  prove  it  of  the  sides  CB,  DE,  I  must  take 
CB  and  DE  for  the  bases  ;  the  reasoning  would  be  the 
same  as  that  I  just  went  through. 

QUERY  DC, 

From  the  ratio  which  you  have  proved  to  exist  between 
the  areas  of  similar  triangles,  can  you  now  find  out  the 
ratio  which  exists  between  the  areas  of  similar  polygons  ? 
(See  Definitions.) 


A.  Yes.  The  areas  of  similar  polygons  are  to  each 
other,  as  the  areas  of  the  squares  constructed  upon  the  cor- 
responding sides.  The  areas  of  the  two  similar  polygons 
ABCDEF,  abcdef  for  instance,  are  to  each  other  as  the 


GEOMETRY.  99 

areas  of  the  squares  constructed  upon  the  sides  AB,  «6, 
or  as  the  areas  of  the  squares  upon  the  sides  BC,  be,  &c. 
For,  by  drawing  in  the  polygon  ABCDEF,  the  diagonals 
AC,  AD,  AE,  and  in  the  polygon  abedef,  the  correspond- 
ing diagonals,  ac,  ad,  ae,  the  triangle  ACB,  is  similar  to 
the  triangle  abc,  the  triangle  ACD,  to  the  triangle  acd, 
&c. ;  because,  if  the  whole  polygons  ABCDEF,  abedef \ 
are  similar,  their  similarly  disposed  parts  must  also  be 
similar  ;  and  the  same  proportion  which  exists  between 
their  parts,  must  necessarily  exist  also  between  the  whole 
polygons;  consequently,  as  the  areas  of  the  triangles 
ABC,  abc,  ACD,  acd,  &c,  are  in  the  ratio  of  the  areas 
of  the  squares  constructed  upon  their  corresponding 
sides,  the  whole  polygons  must  be  in  the  same  ratio, 
which  may  be  expressed  thus : 

Polygon  ABCDEF  :  polygon  abedef 
=  AB  X  AB  :  ab  X  ab- 


RECAPITULATION    OF    THE    TRUTHS    IN    THE    THIRD 
SECTION. 

Ques.  |.  How  do  you  determine  the  length  of  a  line? 

2.  How  do  you  find  out  which  of  two  lines  is  the 
greater  1 

3.  How  can  you  measure  a  surface  ? 

4.  What  do  you  call  the  area  of  a  surface  ? 

5.  If  you  take  one  of  the  sides  of  a  triangle  for  the 
basis,  how  do  you  determine  the  height  of  the  triangle  ? 

6.  How  is  the  height  of  a  parallelogram  determined  ? 
How  that  of  a  rectangle  ?     A  rhombus  ?     A  square  ? 

7.  When  do  you   call  a  triangle  equal  to  a  square? 
to  a  parallelogram  ?  to  a  rectangle,  &>c.  ? 


100  GEOMETRY. 

8.  When  can  you  call  two  geometrical  figures  equal  to 
one  another,  though  these  figures  do  not  coincide  with 
each  other  ? 

9.  Can  you  repeat  the  different  principles  respecting 
the  areas  of  geometrical  figures,  which  you  have  learned 
in  this  section  1 

Ans.  1.  The  area  of  a  rectangle  is  found  by  multiply- 
ing its  basis,  given  in  miles,  rods,  feet,  inches,  &c,  by 
its  height  expressed  in  units  of  the  same  kind. 

2.  The  area  of  a  square  is  found  by  multiplying  one 
of  its  sides  by  itself. 

3.  If  a  parallelogram  stands  on  the  same  basis  as  a 
rectangle,  and  has  its  height  equal  to  the  height  of  that 
rectangle,  the  area  of  the  parallelogram  is  equal  to  the 
area  of  the  rectangle  1 

4.  The  areas  of  all  parallelograms,  which  have  equal 
bases  and  heights,  are  equal  to  one  another. 

5.  Parallelograms  upon  equal  bases,  and  between  the 
same  parallels,  are  equal  to  one  another. 

6.  The  area  of  a  parallelogram  is  found  by  multiplying 
the  basis  given  in  rods,  feet,  inches,  &c,  by  the  height, 
expressed  in  units  of  the  same  kind. 

7.  The  area  of  a  rhombus  or  lozenge  is  found  like 
that  of  a  parallelogram. 

8.  The  areas  of  parallelograms  are  to  each  other,  as 
the  products  obtained  by  multiplying  the  bases  of  the 
parallelograms  by  their  heights. 

9.  Rectangles,  or  parallelograms  which  have  equal 
bases,  are  to  each  other  as  their  heights. 

10.  Rectangles,  or  parallelograms  which  have  equal 
heights,  are  to  each  other  as  their  bases. 

11.  If  two  triangles  stand  on  the  same  basis,  and  have 
equal  heights,  their  areas  are  equal  to  one  another. 


GEOMETRY.  jqj 

12.  Every  triangle  is  half  of  a  parallelogram  upon  equal 
basis  and  of  the  same  height. 

13.  The  area  of  a  triangle  is  half  of  the  area  of  a  par- 
allelogram upon  equal  basis  and  of  the  same  height ;  and, 
therefore,  the  area  of  a  triangle  is  found  by  multiplying 
the  length  of  its  basis  by  its  height,  and  dividing  the 
product  by  2. 

14.  The  areas  of  triangles  upon  the  same  basis,  and 
between  the  same  parallels,  are  equal. 

15.  The  areas  of  triangles  are  to  each  other,  as  the 
products  of  their  bases  by  their  heights. 

16.  The  areas  of  triangles  upon  equal  bases  are  to 
each  other,  as  the  heights  of  the  triangles. 

17.  The  areas  of  triangles,  which  have  equal  heights, 
are  to  each  other,  as  their  bases. 

18.  The  area  of  a  trapezoid  is  found  by  multiplying 
half  the  sum  of  the  two  parallel  sides,  by  their  distance. 

19.  The  area  of  any  rectilinear  figure,  terminated  by 
any  number  of  sides,  is  found  by  dividing  that  figure, 
either  by  diagonals  or  by  any  other  means,  into  triangles,, 
and  then  adding  the  areas  of  these  triangles  together. 

20.  If,  upon  each  of  the  three  sides  of  a  right-angled 
triangle,  a  square  is  constructed,  the  square  upon  the 
hypothenuse  equals,  in  area,  the  two  squares  constructed 
upon  the  two  sides,  which  include  the  right  angle. 

21.  The  bases  of  similar  triangles  are  to  each  other, 
as  the  heights  of  the  triangles. 

22.  The  areas  of  similar  triangles  are  to  each  other, 
as  the  areas  of  the  squares  upon  the  corresponding  sides. 

23.  The  areas  of  similar  polygons  are  to  each  other,  as 
the  squares  constructed  upon  the  corresponding  sides. 

9* 


SECTION  IV. 

OF  THE  PROPERTIES  OF  THE  CIRCLE  * 

QUERY  I. 

In  how  many  points  can  a  straight  line,  CD,  meet  the 
circumference  of  a  circle  1 
A.  In  two  points, 

C : 


M,  N,  only.  For, 
letting  fall,  from  the 
centre  of  the  cir- 
cle, the  perpendicu- 
lar OP  upon  the 
straight  line  CD, 
there  is  but  one  point  in  the  line  CD,  on  each  side  of 
the  perpendicular,  such,  that  a  line,  drawn  from  it  to 
the  point  O  of  the  perpendicular,  has  the  length  of  the 
radius  ON.     (Page  46,  6thly.) 


QUERY  IT. 

In  what  cases  do  the 
circumferences  of  two 
circles  cut  each  other  ? 

A.  When  the  distance, 
OP,  between  their  cen- 
tres, O  and  P,  is  less 
than  the  sum  of  their  radii,  OM,  PM. 

*  Before  entering  on  this  section,  the  teacher  ought  to  recapitu- 
late with  his  pupils  the  definitions  of  a  circle,  of  an  arc,  of  a  chord, 
a  segment,  &c. 


GEOMETRY. 


103 


QUERY  III 

When  do  two  circles 
touch  each  other  exteriorly  ? 

A.  Wlien  the  distance, 
OP,  between  their  centres, 
O  and  P,  is  equal  to  the 
sum  of  their  radii  OM, 
PM. 

QUERY  IV. 

WJicn  do  two  circles  touch  each 
other  interiorly  1 

A.  When  the  distance,  OP,  be- 
tween their  centres,  O  and  P,  is 
equal  to  the  difference  between 
their  radii,  OM  and  PM. 

QUERY  V. 

When  are  the  circumferences  of 
two  circles  parallel  to  each  other  7 

A.  When  they  are  concentric, 
that  is,  when  they  are  described 
from  the  same  point,. C,  as  the 
centre. 

QUERY  VI, 

If  from  the  centre,  C,  of  a  cir- 
cle, a  perpendicular,  CD,  is  let 
fall  upon  a  chord,  AB,  in  that 
circle,  what  relation  do  the  two 
parts,  AD,  BD,  into  which  the 
chord  AB  is  divided,  bear  to  each 
other  1 


104  *  GEOMETRY. 

A.  Tlie  two  parts  AD,  BD,  are  equal  to  each  other ; 
that  is,  the  chord  AB  is  bisected  in  the  point  D. 

Q.  How  can  you  prove  this  1 

A.  By  drawing  the  two  radii  AC,  BC,  the  right-angled 
triangle  ACD  has  the  hypothenuse  AC,  and  the  side  CD, 
equal  to  the  hypothenuse  BC,  and  the  side  CD,  in  the 
right-angled  triangle  BCD,  each  to  each  ;  therefore  these 
two  triangles  are  equal  (page  47)  ;  and  the  side  AD,  in 
the  triangle  ACD,  is  equal  to  the  side  BD  in  the  equal 
triangle  BCD. 

Q.  Wliat  other  truths  can  you  infer  from  the  one  you 
have  just  established  ? 

A.  1.  A  straight  line,  drawn  from  the  centre  of  a  circle 
to  the  middle  of  a  chord,  is  perpendicular  to  that  chord. 

2.  A  perpendicular,  drawn  through  the  middle  of  a 
chord,  passes,  tvhen  sufficiently  far  extended,  through  the 
centre  of  the  circle. 

3.  Two  perpendiculars,  each  drawn  through  the  middle 
of  a  chord  in  the  same  circle,  intersect  each  other  at  the 
centre ;  for  each  of  them  must  go  through  the  centre. 

4.  The  two  angles,  x  and  y,  which  the  radii  AC,  BC, 
drawn  to  the  extremities  of  the  chord  AB,  make  with  the 
perpendicular  CD,  are  equal  to  one  another ;  for  they  are 
opposite  to  the  equal  sides  AD,  BD,  in  the  equal  triangles 
ADC,  BDC. 

QUERY  VH. 

If  the  two  chords,  AD,  AB, 
are  equal  to  each  other,  what  re 
mark  can  you  make  with  regard 
to  the  arcs  AD,  AB,  subtended* 
by  these  chords  ? 

A.   The  two  arcs,  AB,  AD, 

*  The  arcs  AD,  AB,  standing  on  the  chords  AB,  AD,  are  said  to 
be  subtended  by  these  chords. 


GEOMETRY.  105 

subtended  by  the  equal  chords,  AB,  AD,  are  equal  to 
one  another. 

Q.  Why? 

A.  This  follows  from  the  perfect  uniformity  with 
which  a  circle  is  constructed.  For,  if  the  chord  AB  is 
placed  upon  its  equal,  the  chord  AD,  the  arcs,  AB  and 
AD,  must  coincide  with  each  other ;  because  every  point 
in  both  these  arcs  is  at  the  same  distance  from  the  centre, 
C,  of  the  circle. 

Remark.  It  is  to  be  observed  that  each  chord  subtends  two 
arcs,  one  of  which  is  smaller,  and  the  other  greater  than  the  semi- 
circumference,  both  together  completing  the  whole  circumference. 
In  speaking  of  an  arc,  subtended  by  a  chord,  we  always  mean  that 
which  is  smaller  than  the  semi-circumference. 

Q.  TVIiat  other  truths  can  you  infer  from  the  one  you 
have  just  proved  7 

A.  1.  That  equal  arcs  stand  on  equal  chords  ;  for,  by 
placing  one  of  the  equal  arcs  AB,  AD,  upon  the  other, 
the  beginning  and  end  of  the  two  chords  AB,  AD,  and 
therefore  the  whole  chords  themselves,  coincide  with  each 
other. 

2.  The  greater  arc  stands  on  the  greater  chord,  and 
the  greater-  chord  subtends  the  greater  arc.  The  chord 
AF,  for  instance,  is  greater  than  the  chord  AD ;  because 
the  arc  AF,  belonging  to  the  greater  chord  AF,  is  greater 
than  the  arc  AD,  belonging  to  the  smaller  chord  AD. 

3.  Among  all  the  chords,  AD,  AF,  AM,  AN,  AB, 
fyc,  tohich  can  be  drawn  in  a  circle,  the  diameter  AM  is 
the  greatest ;  because  the  greatest  arc,  the  semi-circum- 
ference, stands  on  it. 

Remark.  All  that  has  been  said  of  chords  and  arcs  in  the  same 
circle,  holds  true  also  of  chords  and  arcs  in  equal  circles. 


106  GEOMETRY. 

QUERY  VIII. 

What  relation  do  you  discover 
between  the  angles  ACB,  BCD,  at 
the  centre,  C,  of  a  circle,  and  the 
arcs  AB,  BD,  intercepted  between 
their  legs  ? 

A.  The  angles  ACB,  BCD,  at 
the  centre,  are  to  each  other  in  the 
same  ratio  as  the  arcs  AB,  BD,  of  the  circumference. 

Q.  How  can  you  show  this  ? 

A.  I  divide  the  whole  of  the  arc  AD  successively  into 
smaller  and  smaller  parts,  until  one  of  the  points  of 
division  shall  have  fallen  upon  B.  Then,  it  is  evident 
that  by  drawing  to  the  points  of  division  the  radii  Cm, 
Cn,  Co,  &c,  the  angles  ACB  and  BCD  are  divided  into 
as  many  equal  parts  as  the  arcs  AB,  BD  (for  the  sectors 
ACm,  mCn,  raCB,  &c,  will  all  coincide  with  each  other, 
when  they  are  placed  upon  one  another);  and  therefore 
the  same  ratio  which  exists  between  the  arcs  AB,  BD, 
exists  also  between  the  angles  ACB,  BCD.  In  our  figure, 
we  have  the  ratio  of  the  arc  AB  to  the  arc  BD  as  3  to  6 ; 
and  the  same  ratio  (as  3  to  6)  exists  also  between  the 
angles  ACB  and  BCD  at  the  centre  of  the  circle ;  that 
is,  the  arc  BD  is  as  many  times  greater  than  the  arc  AB, 
as  the  angle  BCD  is  greater  than  the  angle  ACB  (Def. 
of  Geom.  Proportions). 

JVJiat  inference  can  you  draw  from  the  truth  you  have 
just  advanced? 

Ans  1.  If  the  arcs  AB,  BD,  are  equal  to  one  another, 
the  angles  ACB,  BCD,  at  the  centre,  are  also  equal  to 
one  another ;  for  they  are  in  the  same  ratio  as  the  arcs 
AB,  BD  (namely,  then,  in  the  ratio  of  equality). 


GEOMETRY.  107 

2.  If  the  angles  ACS,  BCD,  at  the  centre,  are  equal 
to  one  another,  the  arcs  AB,  BD,  are  also  equal  to  one 
another ;  because  they  are  to  each  other  in  the  same  ratio 
as  the  angles  at  the  centre. 


Remark  1.  It  has  already  been  stated  (note  to  page  12),  that 
angles  are  measured  by  arcs  of  circles,  described  with  any  radius 
between  their  legs.  The  reason  is  now  apparent;  for  the  arcs 
intercepted  between  their  legs  are  in  proportion  to  the  angles  at 
the  centre. 

Remark  2.  If  the  circumference  of  a  circle  is  divided  into  360 
equal  parts,  called  degrees  ;  each  degree  again  into  60  equal  parts, 
called  minutes;  each  minute  again  into  60  equal  parts,  called 
seconds,  &,c;  it  is  easy  to  perceive,  that  the  magnitude  of  an  angle 
does  not  depend  upon  the  length  of  the  arc  intercepted  between 
its  legs  ;  but  merely  upon  the  number  of  degrees,  minutes,  sec- 
onds, &c,  it  measures  of  the  circumference  of  the  circle  of  which 
it  is  a  part. 

Thus,  if  the  angle  BAC  meas- 
ures 3  degrees  by  the  arc  MN, 
it  measures  the  same  number  of 
degrees  by  the  arc  OP,  the  same 
number  of  degrees  by  the  arc  CB, 
&c,  although  the  degrees  them- 
selves vary  in  size. 

Remark  3.  As  the  sum  of  all  the  angles  around  the  same  point 
is  equal  to  4  right  angles  (page  24),  the  sum  of  all  the  angles 
around  the  centre  of  a  circle  is  also  equal  to  4  right  angles ;  there- 
fore the  circumference  of  a  circle  is  the  measure  of  4  right  angles  ; 
the  semi-circumference  that  of  2  right  angles,  and  the  arc  of  a 
quadrant,  that  of  one  right  angle.  If  the  circumference  of  a  circle 
is  divided  into  360  degrees,  90  of  them  are  the  measure  of  1  right 
angle,  180  that  of  2  right  angles,  and  360  that  of  4  right  angles. 


108 


GEOMETRY. 


QUERY  DC. 

If  a  straight  line  is  drawn  per- 
pendicular to  the  extremity  A,  of 
the  radius  AC,  in  how  many 
points  ivill  that  line  meet  the  cir- 
cumference of  the  circle  ? 

A.  In  one  only  {namely,  the 
point  A)  ;  and  therefore  the  line 
DE  is  a  tangent  to  the  circle. 
(See  Def.  page  15.) 

Q.  But  why  can  the  line  ED 
have  no  other  point  common  with  the  circumference  1 

A.  Because  the  perpendicular  AC  is  the  shortest  line 
which  can  be  drawn  from  the  point  C,  the  centre  of  the 
circle,  to  the  straight  line  ED  (page  44) ;  therefore  every 
other  line,  CG,  CF,  CD,  drawn  from  the  centre,  C,  to  the 
straight  line  ED,  will  be  greater  than  the  radius  AC ; 
consequently  every  point  in  the  line  ED,  except  the  point 
A  itself,  is  without  the  circle. 

Q.  And  what  other  truths  can  you  infer  from  the  one 
last  established  ? 

A.  1.  A  radius  or  diameter,  drawn  to  the  point  of 
tangent,  is  perpendicular  to  the  tangent. 

2.  A  line  drawn  through  the  point  of  tangent  perpen- 
dicular to  the  tangent,  passes,  when  sufficiently  far  ex- 
tended, through  the  centre  of  the  circle. 


GEOMETRY.  100 


What  relation  does  the  angle  ACB,  measured  by  the 
arc  AB  bear  to  the  angle  y,  formed  by  the  tangent  BD 
and  the  chord  AB,  which  subtends  the  arc  AB  ? 

A.  The  angle  ACB,  at  the  centre  of  the  circle,  is 
twice  as  great  as  the  angle  y,  formed  by  the  tangent  BD, 
and  the  chord  AB. 

Q.  How  can  you  prove  this  ? 

A.  From  the  centre  of  the  circle,  let  fall  the  perpen- 
dicular CI  upon  the  chord  AB,  and  extend  it  until  it 
meets  the  circumference  in  K.  Then  the  angles  w  and 
z,  and  consequently  the  arcs  AK,  KB,  are  equal  to  one 
another  (page  104,  4thly).  We  have  further  the' triangle 
BIC  right-angled,  and  therefore  the  two  angles  x  and  z, 
together,  equal  to  a  right  angle  (page  34,  Tthly) ;  and 
because  the  tangent  DB  is  perpendicular  to  the  radius 
CB,  the  angles  x  and  y  are  together  also  equal  to  a  right 
angle  ;  therefore  the  angle  z  is  equal  to  the  angle  y 
(Truth  III,  page  21)  :  and  as  the  angle  z  is  half  of  the 
angle  ACB,  the  angle  y  (its  equal)  is  also  half  of  the 
angle  ACB.  *&&* 

Q.  And  what  remark  can  you  make  with  regard  to  the 
arc  BK1 

A.  That  the  arc  BK,  which  measures  the  angle  z,  may 
be  taken  also  for  the  measure  of  the  angle  y  (its  equal); 
10 


HO  GEOMETRY. 

and  as  the  arc  BK  is  half  of  the  arc  AB,  the  angle  y, 
made  by  the  tangent  BD  and  the  chord  AB,  may  likewise 
be  measured  by  half  the  arc  AB. 

Q.  What  do  you  mean  by  saying  that  half  the  arc  AB 
measures  the  angle  y  ? 

A.  That  if  the  arc  AB  is  given  in  degrees,  minutes, 
seconds,  &c,  the  angle  y  measures  half  as  many  degrees, 
minutes,  seconds,  &c,  as  the  arc  AB.  Thus  if  the  arc 
AB  were  12  degrees  and  30  minutes,  the  angle  y  would 
measure  6  degrees  and  15  minutes. 


* 


>>  hat  relation  does  the  angle  w,  formed  by  the  two 
radii  CA,  CF,  bear  to  the  angle  y,  formed  by  the  Uco 
chords  AB,  FB,  if  the  legs  of  both  these  angles  stand  on 
the  extremities  of  the  same  arc  AF  ? 

A.  The  angle  w,  formed  by  the  two  radii  CA,  CF,  is 
twice  as  great  as  the  angle  y,  formed  by  the  two  chords 
AB,  FB. 

Q.  How  can  you  prove  it  ? 

A.  Drawing  in  the  point  B  a  tangent,  DE,  to  the 
circle,  the  angles  x,  y,  z,  being  together  equal  to  two 
right  angles  (duery  4,  Sect.  I.),  will  have  for  their 
measure  half  the  circumference  of  the  circle  (page  107, 
remark  3d).  Now,  the  angle  x,  formed  by  the  tangent 
DB  and  the  chord  AB,  is  measured  by  half  the  arc  AB, 
as  has  been  proved  in  the  last  query ;  and  for  the  same 


GEOMETRY. 


Ill 


reason  is  the  angle  z  measured  by  half  the  arc  BF  ;  and 
therefore  the  remaining  angle  y  is  measured  by  half  the 
arc  AF  ;  because  half  of  the  arc  AF  makes  with  half  of 
the  arcs  AB  and  BF,  half  the  circumference.  But  the 
angle,  w,  at  the  centre  is  measured  by  the  whole  arc  AF; 
therefore  the  angle  w  is  twice  as  great  as  the  angle  y. 

Q.  What  important  truths  can  you  infer  from  the  one 
you  have  just  learned? 

A.  That  every  angle  made  by  two  chords  at  the  cir- 
cumference of  a  circle,  measures  half  as  many  degrees, 
minutes,  seconds,  fyc,  as  the  arc  on  the  extremity  of  which 
these  chords  stand. 

2.  The  angles  x,  y,  z,  at  the 
circumference,  having  their  legs 
standing  on  the  extremities  of  the 
same  arc,  ACB,  are  all  equal  to 
one  another ;  because  each  of 
them  is  measured  by  half  the  arc 
ACB.* 


QUERY  XII. 

If  two  chords,  AH,  CD,  in  the 
same  circle,  are  parallel  to  each 
other,  what  relation  do  the  arcs, 
AC,  BD,  intercepted  by  them,  on 
both  sides  of  the  circumference, 
bear  to  each  other  ? 

A.    The  arcs  AC,  BD,  are  equal  to  each  other. 

Q.  How  can  you  prove  it  ? 

A.  Joining  AD,  the  alternate  angles  x  andy  are  equal 
to  one  another  (Query  10,  Sect.  I.) ;. therefore  the  arcs 


*  The  arc  ACB  is  designated  by  three  letters,  in  order  to  dis- 
tinguish it  from  the  upper  arc  AB. 


112  GEOMETRY. 

AC  and  BD,  measured  by  the  halves  of  these  angles,  are 
also  equal  to  one  another. 

QUERY  XIII. 

If  from  the  same  point,  A, 
without  a  circle,  you  draw  a 
tangent,  AB,  to  the  circle,  and, 
at  the  same  time,  another  line, 
AC,  cutting  the  circle,  what 
relation  exists  between  the  tan- 
gent AB,  and  the  line  AC,  which  cuts  the  circle  ? 

A.  The  tangent  AB  is  a  mean  proportional  (Theory 
©f  Prop.,  page  66),  between  the  whole  line  AC  and  the 
part  AD,  which  is  without  the  circle. 

Q.  How  can  you  prove  it  1 

A.  By  joining  BD  and  BC,  the  triangle  ABD  is  simi- 
lar to  the  whole  triangle  ABC  ;  because  the  angle  at  A 
is  common  to  both  triangles,  and  the  angle  y  in  the  tri- 
angle ABD,  is  equal  to  the  angle  x,  in  the  triangle  ABC 
(both  angles  being  measured  by  half  the  arc  BD*)  ; 
therefore  we  have  the  proportion 

AD  :  AB  =  AB  :  AC, 
where  the  tangent  AB  is  a  mean  proportional  between 
the  whole  line  AC  and  the  part  AD  without  the  circle. 

(The  sides  AD  and  AB,  in  the  triangle  ABD,  are 
opposite  to  the  angles  y  and  z  in  the  same  triangle,  and 
the  sides  AB  and  AC,  ih  the  triangle  ABC,  are  opposite 
to  the  angles  x  and  CBA,  which  are  respectively  equal  to 
the  angles  y  and  z.) 

*  The  angle  x  is  formed  at  the  circumference  by  the  two  chords 
BC  and  DC,  whose  extremities  stand  on  the  arc  BD  (Query  11, 
Sect.  IV.)  ;  and  the  angle  y  is  formed  by  the  tangent  BA,  and  tho 
chord  BD,  which  subtends  the  arc  BD.     (Query  10,  Sect  IV.) 


GEOMETRY. 


113 


QUERY  XIV. 

If  two  chords,  AD,  BC,  cut  each 
other  within  the  circle,  what  relation 
exists  between  the  parts  AE,  ED, 
BE,  EC,  into  which  they  mutually 
divide  each  other  ? 

A.   The  two  parts  AE,  ED,  are 
in  the  inverse  ratio  of  the  two  parts 
BE,  EC ;  that  is,  we  shall  have  the  proportion 
EC  :  EA  —  ED  :  EB,* 

Q.  How  can  you  prove  it  ? 

A.  Joining  AC  and  BD,  the  angle  iv  is  equal  to  the 
angle  z  ;  because  each  of  these  two  angles,  w,  z,  measures 
half  as  many  degrees  as  the  arc  AB  ;  for  the  same  reason 
is  the  angle  x  equal  to  the  angle  y ;  because  each  of  these 
angles  measures  half  as  many  degrees  as  the  arc  CD 
(Query  11,  Sect.  IV.);  and  the  angles  AEC,  BED,  are 
also  equal  to  each  other,  being  opposite  angles  at  the  ver- 
tex (Q,uery  5,  Sect.  I.);  therefore  the  three  angles  of  the 
triangle  AEC  are  equal  to  the  three  angles  of  the  triangle 
BED,  each  to  each  ;  consequently  these  two  triangles 
are  similar  to  one  another ;  and  the  sides  opposite  to  the 
equal  angles,  in  both  triangles,  are  in  the  proportion 

EC  :  EA  —  ED  :  EB 
(EC  and  EA  are  opposite  to  the   angles  x  and  z  in  the 
triangle  AEC  ;  and  ED  and  EB  are  opposite  to  the  an- 


*  The  ratio  ED  to  EB,  is  called  inverse  or  inverted,  because  the 
two  parts  ED,  EB,  are  not  in  direct  proportion  to  the  two  parts 
EC,  EA;  that  is,  the  part  EC  of  the  chord  BC,  is  to  the  part  EA 
of  the  chord  AD,  not  as  the  other  part  EB  of  the  first  chord  BC,  is 
to  the  other  part  ED  of  the  chord  AD,  but  as  the  part  ED  of  the 
second  chord  is  to  the  part  EB  of  the  first  one. 
10* 


114 


GEOMETRY. 


gles  y  and  w,  which  are  equal  to  the  angles  x  and  z,  each 
to  each). 


QUERY  XV. 

If  from  a  point,  A,  without  a  circle, 
two  lines,  AB,  AC,  arc  drawn,  cut- 
ting the  circle ;  what  relation  exists 
between  the  lines  AB,  AC,  and  their 
parts  AD,  AE,  without  the  circle? 

A.  The  whole  lines,  AB,  AC,  are 
to  each  other  in  the  inverse  ratio  of 
their  parts,  AD,  AE,  without  the  cir- 
cle ;  that  is,  we  have  the  proportion, 

AB  :  AC  =  AE  :  AD  (see  the  note  to  page  113). 

Q.  Why  is  this  so  ? 

A.  If  you  join  BE  and  DC,  the  two  triangles  ABE 
and  ADC  are  similar  to  each  other  ;  because  two  angles 
in  the  one  are  equal  to  two  angles  in  the  other,  each  to 
each  (page  73,  1st) ;  the  angle,  at  A,  namely,  is  common 
to  both,  and  the  angles  at  B  and  C  are  equal ;  because 
they  have  the  same  measure  (half  the  arc  DE)  ;  and  as 
in  similar  triangles  the  sides  opposite  to  .the  equal  angles 
are  in  proportion,  we  have 

AB  :  AE  =  AC  :  AD,* 
or,  by  changing  the  order  of  the  mean  terms  (principle  2d 
of  proportion), 

AB  :  AC  r=  AE  :  AD3 
as  above. 


*  The  teacher  will  do  well  to  show  his  pupils  again,  that  the 
sides  AB  and  AE  are  the  corresponding  sides  to  AC  and  AD ; 
because  they  are  opposite  to  the  equal  angles  in  the  triangles. 


GEOMETRY.  115 

Remark  1.  A  regular  polygon  is  a  rectilinear  figure  which  has 
all  its  angles  and  all  its  sides  equal  to  one  another. 

Remark  2.  A  rectilinear  figure  is  said  to  be  inscribed  in  a 
circle,  when  the  vertices  of  all  the  angles  of  that  figure  are  at  the 
circumference  of  the  circle. 

Remark  3.  A  rectilinear  figure  is  said  to  be  circumscribed 
about  a  circle,  when  every  side  of  that  figure  is  a  tangent  to  the 
circle. 

QUERY  XYI. 
If  you  divide  the  circumference 
of  a  circle  into  any  number  of 
equal  parts,  for  instance  into  6 
parts,  and  then  join  the  points 
of  division  by  the  chords  AB, 
BC,  CD,  BE,  EF,  FA,  what 
remark  can  you  make  respecting 
the  rectilinear  figure,  ABCDEF,  which  will  be  inscribed 
in  the  circle  ? 

A.  The  figure  thus  inscribed  in  the  circle  is  a  regular 
polygon. 

Q.  How  can  you  prove  this  ? 

A.  The  circumference  of  the  circle  being  divided  into 
equal  parts,  it  follows  that  the  arcs  AB,  BC,  CD,  &c, 
and  consequently  also  the  chords  AB,  BC,  CD,  &c,  which 
form  the  sides  of  the  inscribed  figure,  are  equal  to  one 
another  (page  105,  1st) ;  and  as  each  of  the  angles  ABC, 
BCD,  CDE,  &,c,  has  its  legs  standing  on  the  whole  cir- 
cumference less  two  of  the  equal  arcs,  into  which  the  cir- 
cumference is  divided,  they  all  measure  the  same  number 
of  degrees,  and  consequently  the  angles  of  the  inscribed 
figure  are  also  equal  to  one  another  ;*  therefore  the  in- 
scribed figure  ABCDEF  is  a  regular  polygon. 

*  The  angle  ABC,  for  instance,  has  its  legs  standing  on  the  whole 
circumference  less  the  two  arcs  AB,  BC ;  and  the  angle  BCD  has 
its  legs  standing  on  the  whole  circumference  less  the  two  equa! 
arcs  BC,  CD,  &c. 


116  GEOMETRY. 

Q.  If  in  this  manner  you  divide  the  circumference  of  a 
circle  into  3,  4,  5,  6,  fyc,  equal  parts,  what  will  be  the 
magnitude  of  each  of  the  arcs  AB,  BC,  CD,  fyc.  ? 

A.  Each  of  the  arcs  AB,  BC,  CD,  fyc,  will  then  be 
"&>  4 >  i" >  i>  4*c->  °f  ^le  whole  circumference,  that  is,  •£,  £, 
-i,  £,  Sfc,  o/"360  degrees,  according  as  the  circumference 
has  been  divided  into  3,  4,  5,  6,  <^c,  parts. 

Q.  And  what  do  you  observe  with  regard  to  the  angles 
x,  y,  z,  &fc,  at  the  centre  of  the  circle,  which  the  radii 
OA,  OB,  OC,  Sfc,  drawn  to  the  points  of  division  A, 
B,  C,  D,  Sfc,  make  with  each  other? 

A.  That  these  angles,  x,  y,  z,  Sfc,  are  all  equal  to  one 
another;  because  they  are  measured  by  the  equal  arcs 
AB,  BC,  CD,  &-c.  They  will  therefore  measure  £,  £,  -£, 
foe,  of  360  degrees,  according  as  the  circumference  of 
the  circle  is  divided  into  3,  4,  5,  &c,  equal  parts. 


QUERY  XVII. 

Can  you  find  the  relation  which  one  of  the  sides  of  a 
regular  inscribed  hexagon  bears  to  the  radius  of  that 
circle?   (See  the  figure  belonging  to  the  last  Query.) 

A.  The  side  of  a  regular  hexagon  inscribed  in  a  circle, 
is  equal  to  the  radius  of  that  circle. 

Q.  Why? 

A.  Because  each  of  the  triangles  ABO,  BCO,  CDO, 
&c,  is  in  the  first  place  isosceles,  two  of  its  sides  being 
radii  of  the  same  circle ;  and  as  each  of  the  angles  x, 
y,  z,  &c,  at  the  centre  of  the  circle,  measures  £  of  360, 
that  is,  60  degrees  (last  Query),  it  follows  that  the  two 
angles  at  the  basis  of  each  of  the  isosceles  triangles  ABO, 
BCO,  CDO,  &,c.  (for  instance,  the  two  angles  w  and  u, 
at  the  basis  of  the  isosceles  triangle  ABO),  measure  to- 
gether 120  degrees ;  because  the  sum  of  the  three  angles 
in  every  triangle  is  equal  to  two  right  angles,  or  180  de- 


GEOMETRY.  J 17 

grees,  and  60  from  180  leave  120  degrees.  Now,  as 
the  two  angles  at  the  basis  of  every  isosceles  triangle  are 
equal  to  each  other  (Query  3,  Sect,  II.) ;  each  of  the 
two  angles  at  the  basis  of  one  of  the  isosceles  triangles 
ABO,  BCO,  DCO,  &c,  will  measure  half  of  120,  that 
is,  60  degrees.  But,  each  of  the  angles  at  the  centre 
measuring  also  60  degrees,  the  three  angles  in  each  of 
the  triangles  ABO,  BCO,  CDO,  &c,  are  equal  to  one 
another ;  and  therefore  these  triangles  are  not  only  isos- 
celes, but  also  equilateral;  consequently  each  of  the  sides 
AB,  BC,  CD,  &c,  of  the  hexagon  is  equal  to  the  radius 
of  the  circle. 

QUERY  XVIH. 

If,  in  a  regular  inscribed 
polygon,  you  draw  from  the  cen- 
tre of  the  circle  the  radii  Oi, 
Ok,  Ol,  Om,  fyc,  perpendicular 
to  the  chords  AB,BC,  CD,  8?c; 
and  at  the  extremities  of  these 
radii,  the  tangents  MN,  NP, 
PQ,  fyc;  what  do  you  observe 

with  regard  to  the  figure  MNPQRS,  circumscribed  about 
the  circle  ? 

A.  The  figure  MNPQRS,  circumscribed  about  the 
circle,  is  a  regular  polygon,  of  the  same  number  of  sides 
as  the  inscribed  polygon,  ABCDEF. 

Q.  How  can  you  prove  this  ? 

A.  The  chords  AB,  BC,  CD,  &c,  are  perpendicular 
to  the  same  radii,  to  which  the  tangents  MN,  NP,  PQ,, 
&c.?%.re  perpendicular  ;  consequently  the  chords  AB,  BC, 
CD,  &c,  are  parallel  to  the  tangents  MN,  NP,  PQ,  &c. 
(for  two  straight  lines,  which  are  both  perpendicular  to  a 
third  line  are  parallel  to  each  other ;  Query  7,  Sect.  I.) ; 


118  GEOMETRY. 

and  therefore  the  triangles  ABO,  BCO,  CDO,  &c,  are 
all  similar  to  the  triangles  MNO,  NPO,  PQO,  &c,  from 
which  they  may  be  considered  as  cut  off,  by  the  lines 
AB,  BC,  CD,  &c.  being  drawn  parallel  to  the  sides  MN, 
NP,  Pa,  &c.  (auery  16,  Sect.  II.)  Now,  as  the  trian- 
gles ABO,  BCO,  CDO,  &c,  are  all  equal  to  one  another, 
the  triangles  MNO,  NPO,  PQO,  &c,  are  all  equal  to 
one  another.  And  therefore  the  circumscribed  figure 
MNPQJIS  is  a  regular  polygon,  similar  to  the  one  in- 
scribed in  the  circle. 

QUERY  XIX. 

It  has  been  proved  (Query  16,  Jg      , g 

Sect.  IV.),  that  a  regular  poly-  -''/z\    i    /^\\ 

gon,    of  any   number  of  sides,        y    ''-\ !/--•''  v 

may  be  inscribed  in  a  circle,  by     &r~fi} LA~ y,  D 

ui 'riding  the  circumference  of  the        \\       /   \       // 

circle  into  as  many  equal  parts  H4 \£'' 

as  the  polygon  shall  have  sides, 

and  then  joining  the  points  of  division  by  straight  lines  : 
can  you  now  prove  the  reverse,  that  is,  that  around  every 
regular  polygon,  a  circle  can  be  drawn  in  such  a  manner, 
that  all  the  vertices  of  the  polygon  shall  be  at  the  circum- 
ference ? 

A.  Yes.  For  I  need  only  bisect  two  adjacent  sides  of 
a  regular  polygon  ;  for  instance,  the  two  sides,  AB,  BC, 
of  the  regular  polygon  ABCDEF ;  and  in  the  points  of 
bisection,  erect  the  two  perpendiculars  g-O,  kO,  which 
will  necessarily  cut  each  other  in  a  point,  O.  Then  it  is 
evident,  that  by  drawing  the  lines  OB,  OC,  OA,  these 
three  lines  are  equal  to  each  other ;  for  the  line  OB  is 
equal  to  OC,  because  the  two  points  B  and  C  are  at  an 
equal  distance  from  the  perpendicular  gO  (page  45, 
5thly) ;  and  for  the  same  reason  is  OB  also  equal  to  OA  j 


GEOMETRY.  UQ 

because  the  points  B  and  A  are  at  an  equal  distance  from 
the  perpendicular  M)  Thus  we  have  in  the  two  trian- 
gles ABO,  BCO,  the  three  sides  in  the  one,  equal  to  the 
three  sides  in  the  other ;  therefore  these  two  triangles  are 
both  isosceles  and  equal  to  each  other. 

Q.  But  of  what  use  is  your  proving  that  the  triangle 
ABO  is  equal  to  the  triangle  BCO? 

A.  It  shows  that  each  of  the  angles  in  the  polygon  is 
bisected  by  one  of  the  lines  OA,  OB,  OC.  For,  in  the 
first  place,  we  have  in  the  two  equal  triangles  BCO, 
ABO,  the  angle  o  equal  to  the  angle  z ;  therefore  the  an- 
gle ABC  is  bisected ;  and  the  angle  o  is  further  equal  to 
the  angle  y,  and  the  angle  z  to  the  angle  v ;  therefore 
the  angles  BCD,  and  FAB,  are  also  bisected.  And 
now  I  can  show  that,  by  drawing  from  the  point  O  the 
lines  OF,  OE,  OD,  to  the  remaining  vertices  F,  E,  D, 
the  whole  polygon  is  divided  into  equal  isosceles  tri- 
angles. Taking,  in  the  first  place,  the  two  triangles, 
AFO  and  ABO,  they  have  two  sides,  OA,  FA,  in  the 
one,  equal  to  two  sides,  OA,  AB,  in  the  other,  each  to 
each ;  and  as  the  angle  FAB  is  bisected  by  the  line  OA, 
the  two  angles  v  and  w  are  also  equal ;  consequently,  the 
two  triangles  AFO,  ABO,  are  equal  to  each  other,  and 
the  angle  u  is  equal  to  the  angle  w  (Query  3,  Sect.  II.). 
In  precisely  the  same  manner  it  may  be  proved  that  the 
triangles  FEO,  EDO,  are  isosceles,  and  equal  to  the  tri- 
angle AFO.  And  as  the  whole  polygon  ABCDEF  is 
thus  divided  into  equal  isosceles  triangles,  the  lines  OA, 
OB,  OC,  OD,  OE,  OF,  are  all  equal  to  one  another ;  and 
therefore,  by  describing  from  the  point  O,  as  a  centre, 
with  a  radius  OA,  a  circle  around  the  polygon  ABCDEF, 
each  of  the  vertices  A,  B,  C,  D,  E,  F,  will  be  in  the  cir- 
cumference of  the  circle. 


120 


GEOMETRY. 


Q.  What  other  important  consequence  follows  from  the 
principle  you  have  just  proved  ? 

A.  That  in  every  regular  polygon  a  circle  may  be 
inscribed  in  such  a  manner,  that  every  side  of  the  polygon 
is  a  tangent  to  the  circle.  For  if,  in  the  regular  polygon 
ABCDEF,  you  describe  with  a  radius  Og  the  circumfer- 
ence of  a  circle,  that  circumference  will  touch  the  middle 
of  the  sides  AB,  BC,  CD,  DE,  EF,  FA,  of  the  polygon 
ABCDEF  ;  because  the  lines  Ok,  Og,  Ol,  &c.  are  all 
equal  to  one  another,  and  will  therefore  be  radii  of  the 
inscribed  circle  ;  and  the  sides  AB,  BC,  CD,  &c,  being 
perpendicular  to  the  radii  Ok,  Og,  Ol,  &c,  will  all  be 
tangents  to  that  circle.    (Page  108.) 


QUERY  XX. 


What  relation  do  you  ob'serve  to  exist  between  two  regu- 
lar polygons,  abcdef,  ABCDEF,  of  the  same  number  of 
sides  ? 

A.    They  are  similar  to  one  another. 

Q.  How  can  you  prove  it '? 

A.  By  describing  a  circle  around  each  of  the  regular 
polygons  abcdef  ABCDEF,  and  drawing  the  radii  Oa, 
Ob,  Oc,  Od,  Oe,  Of,  OA,  OB,  OC,  OD,  OE,  OF,  each 


GEOMETRY.  12! 

of  these  polygons  is  divided  into  as  many  equal  triangles 
as  there  are  sides  in  the  polygon ;  and  as  all  the  angles, 
x,  y,  z,  &lc,  formed  at  the  centre  of  a  regular  polygon, 
are  equal  to  one  another,  I  can  place  the  centre,  O,  of 
the  polygon  abcdef,  upon  the  centre,  O,  of  the  polygon 
ABCDEF,  in  such  a  manner,  that  the  angles  at  the  cen- 
tre shall  all  coincide  with  each  other ;  namely,  so  that 
the  radius  0«  shall  fall  upon  the  radius  OA,  Ob  upon 
OB,  Oc  upon  OC,  &c.  Then,  it  is  evident  that  the 
sides,  ab,  be,  cd,  de,  &c,  of  the  smaller  polygon,  abcdef, 
are  parallel  to  the  sides,  AB,  BC,  CD,  DE,  &,c,  of  the 
greater  polygon,  ABCDEF  ;  for  the  points  a,  b,  c,  d,  e,f 
and  A,  B,  C,  D,  E,  F,  are  in  the  circumferences  of  con- 
centric circles  (Query  5,  Sect.  IV.)  ;  therefore  the  trian- 
gles Oab,  Obc,  Ocd,  &c,  in  the  smaller  polygon,  are  all 
similar  to  the  triangles  OAB,  OBC,  OCD,  &,c,  in  the 
greater  polygon  (Query  16,  Sect.  II.)  ;  consequently  the 
whole  polygon  abcdef  is  similar  to  the  whole  polygon 
ABCDEF. 

Q.  What  other  truths  can  you  infer  from  the  one  you 
have  just  learned? 

A.  1.  The  sums  of  all  the  sides  of  two  regular  polygons 
of  the  same  number  of  sides  are  to  each  other  in  the  same 
ratio  as  the  radii  of  the  inscribed  or  circumscribed  circles. 
For  in  the  two  triangles  ABO  and  abo,  for  instance,  we 
have  the  proportion 

AB  :  ab  ==.  AO  :  ao  ;  that  is, 
the  side  AB  is  as  many  times  greater  than  the  side  ab, 
as  the  radius  AO  is  greater  than  the  radius  ao ;  and  there- 
fore 6  or  any  other  number  of  times  the  side  AB,  is  as 
many  times  greater  than  the  same  number  of  times  the 
side  ab,  as  the  radius  OA  is  greater  than  the  radius  oa; 
that  is,  the  sum  of  all  the  sides  of  the  regular  polygon 
ABCDEF,  is  as  many  times  greater  than  the  sum  of  all 
11 


122  GEOMETRY. 

the  sides  of  the  regular  polygon  abcdef,  as  the  radius  OA 
of  the  circle,  circumscribed  about  the  regular  polygon 
ABCDEF,  is  greater  than  the  radius,  oa,  of  the  circle  cir- 
cumscribed about  the  regular  polygon  abcdef.  In  the 
same  manner  I  can  prove  that  the  sum  of  all  the  sides 
of  the  regular  polygon  ABCDEF,  is  as  many  times  greater 
than  the  sum  of  all  the  sides  of  the  regular  polygon  abcdef, 
as  the  radius  of  the  circle,  inscribed  in  the  regular  poly- 
gon ABCDEF,  is  greater  than  the  radius  of  the  circle 
inscribed  in  the  regular  polygon  abcdef. 

Remark.  The  sum  of  all  the  sides  of  a  geometrical  figure,  that 
is,  a  line  as  long  as  all  its  sides  together,  is  called  the  perimeter  of 
that  figure.  The  above  proportion  may  therefore  be  expressed  in 
shorter  terms  ;  namely,  the  perimeters  of  two  regular  polygons  of 
the  same  number  of  sides,  are  to  each  other  in  the  proportion  of  the 
radii  of  the  inscribed  or  circumscribed  circles. 

2.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides,  are  in  the  same  ratio  as  the  squares 
constructed  upon  the  radii  of  the  inscribed  or  circum- 
scribed circles.  Thus  the  area  of  the  regular  polygon 
ABCDEF,  is  as  many  times  greater  than  the  area  of  the 
regular  polygon  abcdef,  as  the  area  of  the  square  upon 
the  radius  OA  is  greater  than  the  area  of  the  square  upon 
the  radius  oa.  For  the  areas  of  the  similar  triangles 
ABO,  abo,  are  to  each  other  as  the  squares  upon  the 
corresponding  sides  (the  radii  OA,  oa)  ;  therefore,  any 
number  of  times  (in  our  figure  6  timos)  the  areas  of 
these  triangles,  that  is,  the  areas  of  the  regular  polygons 
ABCDEF,  abcdef  themselves,  are  to  each  other  in  the 
same  ratio.  In  the  same  manner  I  can  prove  that  the 
area  of  the  polygon  ABCDEF  is  as  many  times  greater 
than  the  area  of  the  polygon  abcdef  as  the  square  upon 


GEOMETRY.  J  23 

the  circle  inscribed  in  the  regular  polygon  ABCDEF,  is 
greater  than  the  square  upon  the  radius  of  the  circle  in- 
scribed in  the  regular  polygon  abcdef. 


QUERY  XXI. 

From  what  you  have  learn- 
ed of  the  properties  of  regular 
polygons,  can  you  give  a  rule 
for  finding  the  area  of  a  reg- 
ular  polygon  1 

A.  Yes.   Multiply  the  sum 
of  all  the  sides  (the  perimeter) 
by  the  radius  of  the  inscribed 
circle ;  the  product,  divided  by  2,  will  be  the  area  of  the 
regular  polygon. 

Q.  Why? 

A.  Because  every,  regular  polygon,  the  polygon 
ABCDEF,  for  instance,  can  be  divided  into  as  many 
equal  triangles,  as  there  are  sides  in  the  polygon ;  and 
the  area  of  each  of  these  triangles  is  found  by  multiply- 
ing the  basis,  that  is,  one  of  the  sides  of  the  polygon 
by  the  height  (which,  in  every  one  of  these  triangles, 
is  equal  to  the  radius,  om,  of  the  inscribed  circle),  and 
dividing  the  product  by  2 ;  therefore  the  area  of  the 
whole  polygon  ABCDEF  may  at  once  be  found  by  multi- 
plying the  sum  of  all  the  sides  by  the  radius  of  the  in- 
scribed circle,  and  dividing  the  product  by  2.* 


*  Instead  of  multiplying  the  perimeter  by  the  whole  radius,  and 
then  dividing  the  product  by  2,  you  may  at  once  multiply  the 
perimeter  by  half  the  radius,  or  the  radius  by  half  the  perimeter. 


124  GEOMETRY. 


If  you  bisect  each  of  the  arcs  AB,  BC,  CD,  fyc, 
subtended  by  the  sides  AB,  BC,  CD,  fyc,  of  a  regular 
polygon  inscribed  in  a  circle  ;  and  then  to  the  points  of 
division,  m,  n,  o,  p,  q,  r,  draw  the  lines  Am,  mB,  Bn,  nC, 
Co,  Sfc;  what  do  you  observe  with  regard  to  the  regular 
polygon,  AmBnCoDpEqFr,  thus  inscribed  in  the  circle  ? 

A.  The  regular  polygon  AmBnCoDp,  Sfc,  has  twice 
as  many  sides  as  the  regular  polygon  ABCDEF;  for 
the  circumference  of  the  circle  is  now  divided  into  twice 
as  many  equal  parts  as  before.  Thus  if  the  regular 
polygon  ABCDEF  has  6  sides,  the  regular  polygon 
AmBnCoDp,  &>c,  has  12  sides;  and  by  bisecting  again 
the  arcs  Am,  mB,  Bn,  &c,  I  can  inscribe  a  regular  poly- 
gon of  24  sides,  and  so  on,  by  continuing  to  bisect  the 
arcs,  a  regular  polygon  of  48,  96,  192,  &c,  sides. 

Q.  And  what  do  you  observe  with  regard  to  the  arcs 
which  are  subtended  by  the  sides  of  the  polygons, 
ABCDEF  and  AmBwCoDpE^Fr,  inscribed  in  the  circle  1 

A.  The  arcs,  AB,  BC,  CD,  &c,  subtended  by  the 
sides  of  the  regular  polygon  ABCDEF,  first  inscribed  in 


GEOMETRY.    *  125 

the  circle,  stand  farther  off  the  sides  AB,  BC,  CD,  &c, 
than  the  arcs  Am,  mB,  Bra,  6lc,  from  the  sides  Am,  mB, 
Bw,  &.c,  of  the  regular  polygon  of  twice  the  number  of 
sides  ;  consequently,  if  the  arcs  AB,  BC,  CD,  &c,  were 
drawn  out  into  straight  lines,  they  would  differ  more 
from  the  sides  AB,  BC,  CD,  &,c.,  of  the  regular  polygon 
ABCDEF,  first  inscribed  in  the  circle,  than  the  arcs  Am, 
mB,  Bn,  nC,  &c,  would,  in  this  case,  differ  from  the 
sides  Am,  mB,  Bra,  &c,  of  the  regular  polygon  AmBnCo, 
&c,  of  twice  the  number  of  sides. 

Q.  Now,  if,  continuing  to  bisect  the  arcs,  you  inscribe 
regular  polygons  of  24,  48,  96,  192,  &c.  sides,  what 
further  remark  can  you  make  with  regard  to  the  arcs 
subtended  by  the  sides  of  these  polygons? 

A.  These  arcs  differ  less  in  length  from  the  sides 
which  subtend  them,  in  proportion  as  the  polygon  con- 
sists of  a  greater  number  of  sides ;  because,  by  continuing 
to  bisect  the  arcs,  and  thereby  to  increase  the  number  of 
sides  of  the  inscribed  polygons,  the  arcs  subtended  by 
these  sides  grow  nearer  and  nearer  to  the  sides  them- 
selves, and  finally  the  difference  between  them  will  be- 
come imperceptible. 

Q.  And  what  conclusion  can  you  now  draw  respecting 
the  whole  circumference  of  a  circle  ? 

A.  Thai  the  circumference  of  a  circle  differs  very  little 
from  the  sum  of  all  the  sides  of  a  regular  inscribed  poly- 
gon of  a  great  number  of  sides  ;  therefore,  if  the  number 
of  sides  of  the  inscribed  polygon  is  very  great  (several 
thousand  for  instance),  the  polygon  will  differ  so  little 
from  the  circle  itself,  that,  without  perceptible  error,  the 
one  may  be  taken  for  the  other. 
11* 


126  GEOMETRY. 

QUERY  XXIII. 

It  has  been  shown  in  the  last  query,  that  a  circle  may 
be  considered  as  a  regular  polygon  of  a  very  great  num- 
ber of  sides  ;  what  inferences  can  you  now  draw  with 
regard  to  the  circumferences  and  areas  of  circles  ? 


A.  1.  The  circumferences  of  two  circles  are  in  propor- 
tion to  the  radii  of  these  circles  ;  that  is,  a  straight  line 
as  long  as  the  circumference  ofthefrst  circle,  is  as  many 
times  greater  than  a  straight  line  as  long  as  the  circum- 
ference of  the  second  circle,  as  the  radius  AB  ofthefrst 
circle,  is  greater  than  the  radius  ab  of  the  second  circle. 
For  if,  in  each  of  the  two  circles,  a  regular  polygon  of  a 
very  great  number  of  sides  is  inscribed,  the  sums  of  all 
the  sides  of  the  two  polygons  are  to  each  other  in  pro- 
portion to  the  radii,  AB,  ab,  of  the  circumscribed  circles 
(page  122,  1st) ;  and  as  the  difference  between  the  cir- 
cumference of  a  circle  and  the  sum  of  all  the  sides  of  an 
inscribed  polygon  of  a  great  number  of  sides,  is  imper- 
ceptible (last  Query),  we  may  say  that  the  circumferences 
themselves  are  in  the  same  ratio.* 

*  The  teacher  may  give  an  ocular  demonstration  of  this  princi- 
ple, by  taking  two  circles,  cut  out  of  pasteboard  or  wood ;  and 
measuring  their  circumferences  by  passing  a  string  around  them. 
The  measure  of  the  one  will  be  as  many  times  greater  than  the 
measure  of  the  other,  as  the  radius  of  the  first  circle  is  greater  than 
the  radius  of  the  second  circle. 


GEOMETRY.  f$ft 

2.  The  areas  of  two  circles  are  in  proportion  to  the 
squares  constructed  upon  their  radii  ;  that  is,  the  area  of 
the  greater  circle  is  as  many  times  greater  than  the  area 
of  the  smaller  circle,  as  the  area  of  the  square  upon  the 
radius  of  the  greater  circle,  is  greater  than  the  area  of 
the  square  constructed  upon  the  radius  of  the  smaller  circle. 
For  if  in  each  of  these  circles  a  regular  polygon  of  a  great 
number  of  sides  is  inscribed,  the  difference  between  the 
areas  of  the  polygons  and  the  areas  of  the  circles  them- 
selves will  be  imperceptible  ;  and  because  the  areas  of 
two  regular  polygons  of  the  same  number  of  sides  are  in 
the  same  ratio  as  the  areas  of  the  squares  upon  the  radii 
of  the  circles  in  which  they  are  inscribed  (page  122, 
2dly),  the  areas  of  the  circles  themselves  are  in  the  ratio 
of  the  squares  upon  their  radii. 

3.  The  area  of  a  circle  is  found  by  multiplying  the 
circumference  of  the  circle,  given  in  rods,  feet,  inches,  fyc, 
by  half  the  radius,  given  in  units  of  the  same  kind.  Be- 
cause a  circle  differs  so  little  from  a  regular  inscribed 
polygon  of  a  great  number  of  sides,  that  the  area  of  the 
polygon  may,  without  perceptible  error,  be  taken  for  the 
area  of  the  circle.  Now  the  area  of  a  regular  polygon 
inscribed  in  a  circle,  is  found  by  multiplying  the  sum  of 
all  the  sides  by  the  radius  of  the  inscribed  circle,  and 
dividing  the  product  by  2  (page  123)  ;  therefore  the  area 
of  the  circle  itself  is  found  by  multiplying  the  circumfer- 
ence (instead  of  the  sums  of  all  the  sides  of  the  inscribed' 
polygon)  by  the  radius,  and  dividing  the  product  by  2. 
For  it  lias  been  shown  in  the  last  duery,  that  the  sides 
of  a  regular  inscribed  polygon  grow  nearer  and  nearer 
the  circumference  of  the  circumscribed  circle,  in  propor- 
tion as  these  sides  increase  in  number  ;  consequently,  the 
circumference  of  a  circle  inscribed  in  a  regular  polygon 
of  a  great  number  of  sides,  will  also  grow  nearer  and 


128  GEOMETRY. 

nearer  the  circumference  of  the  circumscribed  circle ; 
until  finally  the  two  circumferences  will  differ  so  little 
from  each  other,  that  the  radius  of  the  one  may,  without 
perceptible  error,  be  taken  for  the  radius  of  the  other. 

Remark.  Finding  the  area  of  a  circle  is  sometimes  called 
squaring  the  circle.  The  problem  to  construct  a  rectilinear  fig- 
ure, for  instance  a  rectangle,  whose  area  shall  exactly  equal  the 
area  of  a  given  circle,  is  that  which  is  meant  by  finding  the  quad- 
rature of  the  circle.  For  the  area  of  any  geometrical  figure,  termi- 
nated by  straight  lines  only,  can  easily  be  found  by  the  rule  given 
in  Query  5,  Sect.  Ill;  or, in  other  words,  we  can  always  construct 
a  square,  which  shall  measure  exactly  as  many  square  rods,  feet, 
inches,  &c,  as  a  rectilinear  figure  of  any  number  of  sides. 

Now  it  is  easy  to  show,  that  there  is  nothing  ab9urd  in  the  idea 
of  constructing  a  rectilinear  figure,  for  instance  a  rectangle,  whose 
area  shall  be  equal  to  the  area  of  a  given  circle.  For  let  us  take  a 
semicircle  ABM,  and  let  us  for  a  mo- 
ment imagine  the  diameter  AB  to 
move  parallel  to  itself  between  the 
two  perpendiculars  AI,  BKk  It  is 
evident  that  when  the  diameter  AB  is 
very  near  it*  original  position,  for  in- 
stance in  CD,  the  area  of  the  rectangle  ABCD  is  smaller  than  the 
area  of  the  semicircle  ABM  ;  but  the  diameter  continuing  to  move 
parallel  to  itself  in  the  direction  from  A  to  I,  there  will  be  a  point 
in  the  line  AI,  where  the  area  of  the  rectangle  ABIK  is  greater 
than  the  area  of  the  semicircle  ABM.  Now  as  there  is  a  point  in 
the  line  AI,  below  the  point  I,  in  which  the  area  of  the  rectangle 
ABCD  is  smaller  than  the  area  of  the  semicircle  ABM,  and  as  the 
diameter,  by  continuing  to  move  in  the  same  direction,  makes  in 
different  points  C,  E,  G,  &c,  of  that  same  line,  the  rectangles 
ABCD,  ABEF,  ABGH,  &c,  whose  areas  become  greater  and 
greater,  until  finally  they  become  greater  than  the  area  of  the 
semicircle  itself;  there  mu3t  evidently  be  a  point  in  the  line  AI, 
in  which  a  line  drawn  parallel  to  the  diameter  AB,  makes  with  it 
and  the  perpendiculars  AI,  BK,  a  rectangle,  which,  in  area,  is 
equal  to  the  semicircle  ABM  ;  and  as  there  is  a  rectangle  which, 
in  area,  is  equal  to  the  semicircle  ABM,  by  doubling  it,  we  shall 
have  a  rectangle  which,  in  area,  is  equal  to  the  whole  circle. 


GEOMETRY.  129 

Neither  is  it  difficult  to  find  the  area  of  a  circle  mechanically. 
For  the  arpa  of  a  circle  being  found  by  multiplying  the  circum- 
ference by  the  length  of  the  radius,  and  dividing  the  product  by  2 
(page  127,  3dly),  we  need  only  pass  a  string  around  the  circumfer- 
ence of  a  circle,  and  then  multiply  the  length  of  that  string  by  the 
length  of  the  radius;  the  product  divided  by  2  will  be  the  area  of 
the  circle.  Having  thus  found  the  comparative  length  of  the 
radius  and  circumference  of  one  circle,  we  might  determine  the 
circumference,  and  thereby  the  area  of  any  other  circle,  when 
knowing  its  radius.  For  the  circumferences  of  two  circles  being 
in  proportion  to  the  radii  of  the  two  circles,  we  should  have  three 
terms  of  a  geometrical  proportion  given  ;  viz.  the  radii  of  the  two 
circles,  and  the  circumference  of  the  one ;  from  which  we  might 
easily  find  the  fourth  term  (Theory  of  Proportions,  page  64, 8thly), 
which  would  be  the  circumference  of  the  other  circle. 

But  the  expressions  of  the  circumference  and  area  of  a  circle, 
thus  obtained  by  measurement,  are  never  so  correct  as  is  required 
for  very  nice  and  accurate  mathematical  calculations  ;  we  must 
therefore  resort  to  other  means,  such  as  geometry  itself  furnishes, 
to  calculate  the  ratio  of  the  radius  or  diameter  to  the  circumfer- 
ence ;  and  herein  consists  the  difficulty  of  the  quadrature  of  the 
circle.  For  if  the  ratio  of  the  radius  to  the  circumference  is  once 
determined,  we  can  easily  find  the  circumference  of  any  circle, 
when  its  radius  is  given  ;  and  knowing  the  circumference  and  the 
radius,  we  can  find  the  area  of  the  circle. 

To  calculate  the  ratio  of  the  diameter  to  the  circumferencey 
mathematicians  have  compared  the  circumference  of  a  circle  to 
the  sum  of  all  the  sides  of  a  regular  inscribed  polygon  of  a  great 
number  of  sides;  for  it  has  been  shown  (page  125),  that  the  cir- 
cumference of  a  circle  differs  very  little  from  the  sum  of  all  the 
sides  of  such  a  polygon. 

For  this  purpose  they  took  a  regular  inscribed  hexagon,  each 
of  the  sides  of  which  is  equal  to  the  radius  of  the  circumscribed 
circle.  (Query  17,  Sect.  IV.)  For  the  sake  of  convenience  they 
supposed  the  diameter  of  the  circle  equal  to  unity ;  the  radius  and 
therefore  the  side  of  a  regular  inscribed  hexagon  is  then  ^,  and 
the  sum  of  all  the  sides  (6  times  ^)  equal  to  3.  This  is  the  first 
approximation  to  the  circumference  of  a  circle. 

From  the  side  of  a  regular  inscribed  hexagon,  it  is  easy  to  find 
that  of  a  regular  inscribed  polygon  of  12  sides.     Supposing,  for 


130  GEOMETRY. 

instance,  the  chord  CD  to  be  the  side  of  a  regular  inscribed  hex- 
agon, by  bisecting  the  arc  CD  in  B,  the  chords  BC,  BD,  will  be 
two  sides  of  a  regular  inscribed  poly- 
gon of  12  sides,  the  length  of  which 
can  easily  be  calculated  when  the 
chord  CD  and  the  radius  AC  are  once 
known.  For  the  radius  AB  which 
bisects  the  arc  CD,  makes  the  angles 
x  and  y,  which  are  measured  by  the 
arcs  DB,  BC,  equal  to  each  other ; 
and  therefore  AE  is  perpendicular  to 
the  chord  DC,  and  bisects  it  in  E. 
(Page  104,  4thly.) 

Now  the  radius,  AC  and  EC  (half 
of  CD),  being  known,  the  hypothenuse  and  one  of  the  sides  of  the 
right-angled  triangle  AEC  are  given,  whence  it  is  easy  to  find  the 
other  side  AE,  by  the  rule  given  in  the  remark,  page  95.  Thus  if 
the  radius  is  supposed  to  be  £,  the  side  CD  of  the  inscribed  hexa- 
gon is  also  equal  to  £ ;  and  EC  (half  of  CD)  is  \.  Taking  the 
square  of  ^  from  that  of  £,  and  extracting  the  square  root  of  the 
remainder,  we  obtain  the  length  of  the  side  AE,  which,  subtracted 
from  the  radius  AB,  leaves  the  length  of  BE.  Now  we  can  find 
the  side  BC  in  the  right-angled  triangle  BCE,  by  extracting  the 
square  root  of  the  sum  of  the  squares  of  BE  and  EC  (see  the  re- 
mark, page  95)  ;  and  one  of  the  sides  of  the  regular  inscribed  polygon 
of  12  sides  being  once  determined,  we  need  only  multiply  it  by  12, 
in  order  to  obtain  the  sum  of  all  its  sides,  which  is  the  second 
approximation  to  the  circumference  of  the  circle.  In  precisely  the 
same  manner  can  the  side,  and  consequently  also  the  sum  of  all 
the  sides  of  a  regular  inscribed  polygon  of  24  sides  be  obtained, 
when  that  of  a  regular  inscribed  polygon  of  12  sides  is  once 
known ;  which  is  the  third  approximation  to  the  circumference. 
Thus  we  might  go  on  finding  the  sum  of  all  the  sides  of  a  regular 
inscribed  polygon  of  48,  96,  192,  &c,  sides,  until  the  inscribed 
polygon  should  consist  of  several  thousand  sides:  the  sum  of  all  the 
sides  would  then  differ  so  little  from  the  circumference  of  the  cir- 
cle, that,  without  perceptible  error,  we  might  take  the  one  for  the 
other.  »>r 

In  this  manner  the  approximation  to  the  circumference  of  the 
circle  has  been  carried  further  than  is  ever  required  in  the  minutest 
and  most  accurate  mathematical  calculations. 


GEOMETRY.  13] 

The  beginning  of  this  extremely  tedious  calculation  gives  the 
following  results : 


Parts  of  the  cir- 
cumference. 

Sides  of  the  inscribed 
polygon. 

Sum   of  all  the  sides  of  th« 
inscribed  polygon. 

6 

0,5 

3 

12 

0,258819 

3,105828 

24 

0,130526 

3,132628 

48 

0,065403 

3,139348 

96 

0,032719 

3,141033 

192 

0,016361 

3,141446 

It  is  not  necessary  to  carry  this  calculation  any  further,  since 
analysis  furnishes  us  with  means  to  obtain  the  same  results  in  a 
much  easier  manner. 

In  nearly  the  same  manner  has  Loudolph  van  Ceulen 
found  the  ratio  of  the  diameter  to  the  circumference  of  a  circle 
to  32  decimals.  (See  his  *  Arithmetische  en  Geom.  Fundamenten, 
page  163.  Leiden.  1616 ;'  also  his  work  «  De  Circulo  et  Adscriptis, 
c.  10.     Leiden.  1619.'*) 

Archimedes  found  the  ratio  of  the  diameter  to  the  circumfer- 
ence as  near  7  to  22. 

Franciscus  Vieta  found  it  as  1  to  3,1415926535. 
Adrianus  Romanus  added  the  following  decimals 

89793. 
Lotjdolph  van  Ceulen  added  further 

23846264338327950288. 
Sharp  added  again 

41971693993751058209749445923078, 
To  which  Machin  further  added 

164062862089986280348253421170679, 
And  lastly  Lagny  increased  them  by 

821480865132823066470938446. 
In  a  manuscript  in  the  library  at  Oxford,  this  number  is  still 
further  extended  by  29  decimals,  namely, 

460955051822317253594081284802. 

*  The  first  work  which  Loudolph  van  Ceulen-  published  on 
this  subject,  bears  the  title  '  Van  den  Circkel,  daer  in  Gheleert 
wird  te  vinden  de  naeste  proportie  des  Cirkels-Diameter  tegen 
synen  Omloop.  Leiden,  d.  20  Sept.  1596.'  The  work  is  dedicated 
to  Prince  Moriz  of  Orange 


13£  GEOMETRY. 

So  that  the  most  accurate  ratio  of  the  diameter  to  the  circum- 
ference is  at  present  as 

I  to  3,1415926535897932384626433832795028841971693993751 
0582097494459230781640628620899862803482534211706 
7982148086513282306647093844646095505182231725359 
4081284802. 

The  last  ratio  is  so  near  the  truth,  that  in  a  circle,  whose  diame- 
ter is  one  hundred  million  times  greater  than  that  of  the  sun,  the 
error  would  not  amount  to  the  one  hundred  millionth  part  of  the 
breadth  of  a  hair. 

In  general,  when  the  calculations  need  not  be  very  minute  and 
accurate,  7  decimals  will  suffice.  Thus  we  may  consider  the  ratio 
of  the  diameter  to  the  circumference  to  be 

as  1  to  3,1415926  ;  that  is, 
if  the  diameter  of  a  circle  is  1,  its  circumference  is  3,1415926;* 
consequently  if  the  diameter  is  2, -or  the  radius  1,  the  circum- 
ference will  be  twice  3,1415926,  equal  to  6,2831852.  Dividing 
this  number  by  360,  we  obtain  the  length  of  a  degree  ;  dividing 
the  length  of  a  degree  by  60,  we  obtain  the  length  of  a  minute  ; 
and  that  again  divided  by  60,  gives  the  length  of  a  second,  and  so 
on.     In  this  manner  we  obtain  the  length  of 

1  degree  equal  to  0,0174533t 
1  minute  "  "  0,0002909 
1  second  "  "  0,0000048 
1  third         "       "   0,0000001 

Having  once  determined  the  circumference  of  the  circle  whose 
radius  is  1,  we  can  easily  find  the  circumference  of  any  circle, 
when  its  radius  is  given  ;  for  we  need  only  multiply  the  number 
6,2831852  (that  is,  the  circumference  of  a  circle  whose  radius  is  1), 
by  the  radius  of  the  circle  whose  circumference  is  to  be  found; 
the  product  will  be  the  circumference  sought.  Thus  if  it  is  required 
to  find  the  circumference  of  a  circle  whose  radius  is  6  inches,  we 
need  only  multiply  the  number  6,2831852  by  6 ;  the  product 
37,6991112  is  the  circumference  of  that  circle. 

If  it  be  required  to  find  the  length  of  an  arc  of  a  given  number 

*  The  number  3,1415926  is  sometimes  represented  by  the  Greek 
letter  n.  Thus  the  circumference  of  a  circle,  whose  radius  is  1, 
may  be  represented  by  2*. 

t  The  last  figure  in  these  expressions  has  been  corrected, 


GEOMETRY.  133 

of  degrees,  minutes,  seconds,  &c,  in  a  given  circle,  we  need  only 

multiply 

the  degrees  by  0,0174533 
the  minutes  by  0,0002909 
the  seconds  by  0,0000048,  &c  ; 
the  different  products  added  together  give  the  length  of  an  arc  of 
the  same  number  of  degrees,  minutes,  seconds,  &c,  in  the  circle 
whose  radius  is  1  ;  and  multiplying  this  product  by  the  radius  of 
the  given  circle,  we  shall  have  the  length  of  the  arc  sought.     If, 
for  instance,  it  be  required  to  find  the  length  of  an  arc  of  6  degrees 
and  2  minutes,  in  a  circle  whose  radius  is  5  inches,  we  in  the  first 
place  multiply  0,0174533  by  6,  and 

0,0002909  by  2 ;  the  products  of 

these  multiplications,  0,1047198  } 

.   _  nnn~0i0  >    added  together, 
and  0,000o818  $  b 

give  0,1053016,  which  is  the  length  of  an  arc  of 

6  degrees  and  2  minutes  in  the  circle  whose  radius  is  1,  and  this 

last  product  (0,1053016),  multiplied  by  5,  gives  0,5265080,  which  is 

the  length  of  an  arc  of  the  same  number  of  degrees  and  minutes  of 

a  circle  whose  radius  is  5  inches. 

Now  that  we  are  able  to  find  the  circumference,  or  an  arc  of  the 
circumference  of  any  circle,  when  knowing  its  radius,  nothing 
can  be  easier  than  to  calculate  the  area  of  a  circle,  of  a  sector,  a 
segment,  &c. 

The  area  of  a  circle  being  found  by  multiplying  the  circumfer- 
ence by  half  the  radius,  or  by  multiplying  half  the  circumference 
by  the  whole  radius  (page  127,  3dly),  we  need  only  take  the  number 
3,1415926,  which  is  half  the  circumference  of  the  circle  whose 
radius  is  1,  and  multiply  it  by  the  radius  of  the  given  circle  ;  the 
product  will  be  half  the  circumference  of  the  given  circle,  which 
multiplied  again  by  the  radius,  gives  us  the  area  of  it.  Thus  if  it 
is  required  to  find  the  area  of  a  circle  whose  radius  is  5  inches,  we 
multiply  the  number  3,1415926  twicein  succession  by  5,  that  is, 
we  multiply  it  by  the  square  of  5  ;*  the  product  78,5398150  is  the 
area  sought.     Hence  follows  the  general  rule  : 

In  order  to  find  the  area  of  a  circle,  multiply  the  number 
3,1415926  by  the  square  of  the  radius. 

*  Multiplying  a  number  twice  in  succession  by  5,  is  the  same  as 
multiplying  that  number  by  25  ;  which  is  the  square  of  5  ;  because 
5  times  5  are  25. 

12  .n«* 


134  GEOMETRY. 

If  the  radius  is  given  in  rods,  the  answer  will  be  square  rods ;  if 
given  in  feet,  the  answer  will  be  square  feet,  if  in  seconds,  square 
seconds,  and  so  on.  The  area  of  a  semicircle  is  found  by  dividing 
the  area  of  the  whole  circle  by  2.  In  the  same  manner  we  find 
the  area  of  a  quadrant  by  dividing  the  area  of  the  whole  circle 
by  4,  &c. 

The  area  of  a  sector  BCAD  is  found  by 
multiplying  the  length  of  the  arc  CD  A  by 
half  the  radius,  or  we  may  first  find  what 
part  of  the  circumference  the  arc  CDA  is ; 
whether  a  third,  a  fourth,  a  fifth,  &c,  and 
then  divide  the  area  of  the  whole  circle 
whose  radius  is  BC,  by  3,  4,  5,  &.c,  according 
as  the  arc  CDA  is  |,  \,  &c,  of  the  whole 
circumference.  If  we  are  to  find  (he  area 
of  the  segment  CDA,  we  must  first  find  the 
area  of  the  sector  BCDA ;  then  the  area  of  the  triangle  ABC  ; 
which,  subtracted  from  the  area  of  the  sector  BCDA,  will  leave 
the  area  of  the  segment  CDA. 


RECAPITULATION    OF  THE    TRUTHS  CONTAINED  IN 
THE  FOURTH  SECTION. 

Q.  Can  you  now  repeat  the  different  relations  which 
exist  between  the  different  parts  of  a  circle  and  the 
straight  lines,  which  cut  or  touch  the  circumference  ? 

A.  1.  A  straight  line  can  touch  the  circumference 
only  in  one  point. 

2.  When  the  distance  between  the  centres  of  two  cir- 
cles is  less  than  the  sum  of  their  radii,  the  two  circles 
cut  each  other. 

3.  When  the  distance  between  the  centres  of  two  cir- 
cles is  equal  to  the  sum  of  their  radii,  the  two  circles 
touch  each  other  exteriorly. 

4.  When  the  distance   between   the   centres  of  two 


GEOMETRY.  135 

circles  is  equal  to  the  difference  between  their  radii,  the 
two  circles  touch  each  other  interiorly. 

5.  When  two  circles  are  concentric,  that  is,  when  they 
are  both  described  from  the  same  point  as  a  centre,  the 
circumferences  of  the  two  circles  are  parallel  to  each 
other. 

6.  A  perpendicular,  let  fall  from  the  centre  of  a  circle, 
upon  one  of  the  chords  in  that  circle,  divides  that  chord 
into  two  equal  parts. 

7.  A  straight  line,  drawn  from  the  centre  of  a  circle 
to  the  middle  of  a  chord,  is  perpendicular  to  that  chord. 

8.  A  perpendicular,  drawn  through  the  middle  of  a 
chord,  passes,  when  sufficiently  far  extended,  through  the 
centre  of  the  circle. 

9.  Two  perpendiculars,  each  drawn  through  the  mid- 
dle of  a  chord  in  the  same  circle,  intersect  each  other  at 

the  centre. 

i 

10.  The  two  angles  which  two  radii,  drawn  to  the 
extremities  of  a  chord,  make  with  the  perpendicular  let 
fall  from  the  centre  of  the  circle  to  that  chord,  are  equal 
to  one  another. 

11.  If  two  chords  in  the  same  circle,  or  in  equal  cir- 
cles, are  equal  to  one  another,  the  arcs  subtended  by 
them  are  also  equal ;  and  the  reverse  is  also  true ;  that 
is,  if  the  arcs  are  equal  to  one  another,  the  chords  which 
subtend  them  are  also  equal. 

12.  The  greater  arc  stands  on  the  greater  chord,  and 
the  greater  chord  subtends  the  greater  arc. 

13.  The  angles  at  the  centre  of  a  circle  are  to  each 
other  in  the  same  ratio,  as  the  arcs  of  the  circumference 
intercepted  by  their  legs. 

14.  If  two  angles  at  the  centre  of  a  circle  are  equal  to 
one  another,  the  arcs  of  the  circumference,  intercepted 
by  their  legs,  are  also  equal ;  and  the  reverse  is  also  true ; 


136  GEOMETRY' 

that  is,  if  the  two  arcs  intercepted  by  the  legs  of  the  two 
angles  at  the  centre  of  a  circle,  are  equal  to  one  another, 
these  angles  are  also  equal. 

15.  Angles  are  measured  by  arcs  of  circles,  described 
with  any  radius  between  their  legs.  The  circumference 
is,  for  this  purpose,  divided  into  360  equal  parts,  called 
degrees;  each  degree  into  60  equal  parts,  called  min- 
utes ;  each  minute,  again,  into  60  equal  parts,  called  sec- 
onds, &c. 

16.  The  magnitude  of  an  angle  does  not  depend  on 
the  length  of  the  arc  intercepted  by  its  legs;  but  merely 
on  the  number  of  degrees,  minutes,  seconds,  &c,  it 
measures  of  the  circumference. 

17.  The  circumference  of  a  circle  is  the  measure  of  4 
right  angles;  the  semi-circumference  that  of  2  right  an- 
gles ;  and  a  quadrant  that  of  1  right  angle. 

1$.  A  straight  line  drawn  at  the  extremity  of  the 
diameter  or  radius,  perpendicular  to  it,  touches  the  cir- 
cumference only  in  one  point,  and  is  therefore  a  tangent 
to  the  circle. 

19.  A  radius  or  diameter  drawn  to  the  point  of  tangent, 
is  perpendicular  to  the  tangent. 

20.  A  line  drawn  through  the  point  of  tangent,  per- 
pendicular to  the  tangent,  passes,  when  sufficiently  far 
extended,  through  the  centre  of  the  circle. 

21.  The  angle,  formed  by  a  tangent  and  a  chord,  is 
half  of  the  angle  at  the  centre,  which  is  measured  by  the 
arc  subtended  by  that  chord  ;  therefore  the  angle,  formed 
by  the  tangent  and  the  chord,  measures  half  as  many 
degrees,  minutes,  seconds,  &c,  as  the  angle  at  the 
centre. 

22.  The  angle  which  two  chords  make  at  the  cir- 
cumference of  a  circle,  is  half  of  the  angle  made  by 
two  radii   at  the  centre,  having  its  legs  stand  on  the 


GEOMETRY.  137 

extremities  of  the  same  arc  ;  therefore  every  angle,  made 
by  two  chords  at  the  circumference  of  a  circle,  measures 
half  as  many  degrees,  minutes,  seconds,  &c,  as  the  arc 
intercepted  by  its  legs. 

23.  If  several  angles  at  the  circumference  have  their 
legs  stand  on  the  extremities  of  the  same  arc  these 
angles  are  all  equal  to  one  another. 

24.  Parallel  chords  intercept  equal  arcs  of  the  circum- 
ference. 

25.  If,  from  a  point  without  the  circle,  you  draw  a  tan- 
gent to  the  circle,  and,  at  the  same  time,  a  straight  line 
cutting  the  circle,  the  tangent  is  a  mean  proportional  be- 
tween that  whole  line,  and  that  part  of  it  which  is  without 
the  circle. 

26.  If  a  chord  cuts  another  loitldn  the  circle,  the  two 
parts,  into  which  the  one  is  divided,  are  in  the  inverse 
ratio  of  the  two  parts,  into  which  the  other  is  divided. 

27.  If,  from  a  point  without  a  circle,  two  straight  lines 
are  drawn,  cutting  the  circle,  these  lines  are  to  each 
other  in  the  inverse  ratio  of  their  parts  without  the  circle. 

28.  If  the  circumference  of  a  circle  is  divided  into 
3,  4,  5,  &-c,  equal  parts,  and  then  the  points  of  division 
are  joined  by  straight  lines,  the  rectilinear  figure,  thus 
inscribed  in  the  circle,  is  a  regular  polygon  of  as  many 
sides,  as  there  are  parts  into  which  the  circumference  is 
divided. 

29.  If,  from  the  centre  of  a  regular  polygon,  inscribed 
in  a  circle,  radii  are  drawn  to  all  the  vertices  at  the  cir- 
cumference, the  angles  which  these  radii  make  with  each 
other  at  the  centre,  are  all  equal  to  one  another. 

30.  The  side  of  a  regular  hexagon  inscribed  in  a  circle, 
is  equal  to  the  radius  of  the  circle. 

31.  If,  from  the  centre  of  a  circle,  radii  are  drawn, 

bisecting  the  sides  of  a  regular  inscribed  polygon,  and 
to  * 


138  GEOMETRY. 

then,  at  the  extremities  of  these  radii,  tangents  are  drawn 
to  the  circle,  these  tangents  form  with  each  other  a  regu- 
lar ci?'cwnscribed  polygon  of  the  same  number  of  sides  as 
the  regular  inscribed  polygon. 

32.  Around  every  regular  polygon  a  circle  can  be 
drawn  in  such  a  manner,  that  all  the  vertices  of  the  poly- 
gon shall  be  at  the  circumference  of  the  circle. 

33.  Two  regular  polygons  of  the  same  number  of  sides 
are  similar  figures. 

34.  The  sums  of  all  the  sides  of  two  regular  polygons 
of  the  same  number  of  sides,  are  to  each  other  in  the 
same  ratio,  as  the  radii  of  the  inscribed  or  circumscribed 
circles. 

35.  The  areas  of  two  regular  polygons  of  the  same 
number  of  sides,  are  to  each  other  as  the  areas  of  the 
squares  constructed  upon  the  radii  of  the  inscribed  or 
circumscribed  circles. 

36.  The  area  of  a  regular  polygon  is  found  by  multiply- 
ing the  sum  of  all  its  sides  by  the  radius  of  the  inscribed 
circle,  and  dividing  the  product  by  2 ;  or  we  may  at  once 
multiply  half  the  sum  of  all  the  sides  by  the  radius  of  the 
inscribed  circle,  or  half  that  radius  by  the  sum  of  all  the 
sides. 

37.  If  the  arcs  subtended  by  the  sides  of  a  regular 
polygon,  inscribed  in  a  circle,  are  bisected,  and  chords 
drawn  from  the  extremities  of  these  arcs  to  the  points  of 
division,  the  new  figure  thus  inscribed  in  the  circle,  is  a 
regular  polygon  of  twice  the  number  of  sides  as  the  one 
first  inscribed. 

38.  The  circumference  of  a  circle  differs  so  little  from 
the  sum  of  all  the  sides  of  a  regular  inscribed  polygon  of 
a  great  number  of  sides,  that,  without  perceptible  error, 
the  one  may  be  taken  for  the  other. 

39.  The  circumferences  of  two  circles  are  in  proportion 


GEOMETRY.  *      139 

to  the  radii  of  these  circles ;  that  is,  a  straight  line,  as 
long  as  the  circumference  of  the  first  circle,  is  as  many 
times  greater  than  a  straight  line  as  long  as  the  circum- 
ference of  the  second  circle,  as  the  radius  of  the  one  is 
greater  than  the  radius  of  the  other. 

40.  The  areas  of  two  circles  are  in  proportion  to  the 
squares  constructed  upon  their  radii ;  that  is,  the  area  of 
the  greater  circle  is  as  many  times  greater  than  the  area 
of  the  smaller  circle,  as  the  area  of  the  square  upon  the 
radius  of  the  one  is  greater  than  the  area  of  the  square 
upon  the  radius  of  the  other. 

41.  The  area  of  a  circle  is  found  by  multiplying  the 
circumference,  given  in  rods,  feet,  inches,  &c,  by  half 
the  radius,  given  in  units  of  the  same  kind. 

42.  The  circumference  of  a  circle,  whose  radius  is  1, 
is  equal  to  the  number  6,2831852 ;  and  the  circumference 
of  any  other  circle  is  found  by  multiplying  the  number 
6,2831852  by  the  length  of  the  radius. 

43.  The  length  of  1  degree  in  a  circle,  whose  radius 
is  1,  is  equal  to  the  number  0,0174533 

The  length  of  1  minute  0,0002909 

"         "       "  1  second  0,0000048 

"         "       "  1  third  0,0000001 

44.  The  length  of  an  arc,  given  in  degrees,  minutes, 
seconds,  &c,  is  found  by  multiplying  the  degrees  by 
0,0174533,  the  minutes  by  0,0002909,  the  seconds  by 
0,0000048,  &c,  then  adding  these  products  together, 
and  multiplying  their  sum  by  the  radius  of  the  circle. 

45.  The  area  of  a  circle,  whose  radius  is  1,  is  equal  to 
3,1415926  square  units;  and  the  area  of  any  other  circle 
is  found  by  multiplying  the  number  3,1415926  by  the 
square  of  the  radius. 

46.  The  area  of  a  semicircle  is  found  by  dividing  the 
area  of  the  whole  circle  bv  2. 


140   ♦  GEOMETRY. 

47.  The  area  of  a  quadrant  is  found  by  dividing  the 
area  of  the  whole  circle  by  4. 

48.  The  area  of  a  sector  is  found  by  multiplying  the 
length  of  the  arc  by  half  the  radius. 

49.  In  order  to  find  the  area  of  a  segment,  we  first 
draw  two  radii  to  the  extremities  of  the  arc  of  that  seg- 
ment ;  then  calculate  the  area  of  the  sector,  formed  by 
the  two  radii  and  that  arc,  and  subtract  from  it  the  area 
of  the  triangle  formed  by  the  two  radii  and  the  chord  of 
the  segment:  the  remainder  is  the  area  of  the  segment. 


SECTION    V, 


£*>- 


APPLICATION    OF   THE    FOREGOING  PRINCIPLES  TO 
THE   SOLUTION  OF   GEOMETRICAL  PROBLEMS- 

PART   I. 

Problems  relative  to  the  drawing  and  division  of 
lines  and  angles. 

Problem  I.  To  construct  an  equilateral  triangle  upon 
a  given  straight  line,  AB. 

Solution.     Let  AB  be  the 
given  straight  line. 

1.  From  the  point  A,  as  a 
centre,    with   the    radius   AB, 
describe  an  arc  of  a  circle,  and 
from    the   point   B,    with    the       £  r 
same  radius,  AB,  another  arc  cutting  the  first. 

2.  From  the  point  of  intersection,  C,  draw  the  lines 
AC,  BC ;  the  triangle  ABC  will  be  equilateral. 

Demonstration.  The  three  sides,  AB,  AC,  BC,  of  the  trian- 
gle ABC,  are  all  equal  to  each  other ;  because  they  are  radii  of 
equal  circles. 

Remark.  In  a  similar  manner  can  an  isosceles  triangle  be  con- 
structed upon  a  given  basis. 

a 


142 


GEOMETRY. 


Problem  II.  From  a  given  point  in  a  straight  line,  to 
erect  a  perpendicular  upon  that  line. 


M 


B      J) 


N 


I.  Solution.  Let  MN  be  the  given  straight  line,  and 
D  the  point  in  which  the  perpendicular  is  to  be  erected* 

1.  Take  any  distance,  BD,  on  one  side  of  the  point  D, 
and  make  DA  equal  to  it. 

2.  From  the  point  B,  with  any  radius  greater  than 
BD,  describe  an  arc  of  a  circle,  and  from  the  point  A, 
with  the  same  radius,  another  arc,  cutting  the  first. 

3.  Through  the  point  of  intersection,  C,  and  the  point 
D,  draw  a  straight  line,  CD,  which  will  be  perpendiculai 
to  the  line  MN. 

Demojv.  The  three  sides  of  the  triangle  BCD,  are  equal  to  th6 
three  sides  of  the  triangle  ACD,  each  to  each,  viz.  • 

the  side  BC  equal  to  AC 
m      «     uj)     «     «  da 

«      «     CD     "       "CD; 

therefore  the  three  angles  in  the  triangle  BCD  are  also  equal  to 
the  three  angles  of  the  triangle  ADC,  each  to  each  (page  40)  ; 
and  the  angle  x  opposite  to  the  side  BC  in  the  triangle  BCD,  is 
equal  to  the  angle  y  opposite  to  the  equal  side  AC  in  the  triangle 
ACD ;  and  as  the  two  adjacent  angles,  which  the  line  CD  makes 
with  the  line  MN,  are  equal  to  one  another,  the  line  CD  is  per- 
pendicular to  MN.     (Definitions  of  perpendicular  lines,  page  12.) 


143 


II.  Solution.  Let  MN  be  the  given  straight  line, 
and  A  the  point  in  which  the  perpendicular  is  to  be 
drawn  to  it. 

1.  From  a  point,  O,  as  a  centre,  with  a  radius,  OA, 
greater  than  the  distance  O  from  the  straight  line  MN, 
describe  the  circumference  of  a  circle. 

2.  Through  the  point  B  and  the  centre  O,  of  the  cir- 
cle, draw  the  diameter  BC. 

3.  Through  C  and  A  draw  a  straight  line,  which  will 
be  perpendicular  to  the  line  MN. 

Demon.  The  angle  BAC,  at  the  circumference,  measures  half 
as  many  degrees  as  the  arc  BPC  intercepted  by  its  legs  (page  111, 
1st).  But  the  arc  BPC  is  a  semi-circumference  ;  therefore  the  angle 
BAC,  measures  a  quadrant ;  consequently  the  angle  BAC  is  a  right 
angle  (page  107,  Remark  3),  and  the  line  AC  is  perpendicular  to 
MN. 


Problem  III.     To  bisect  a  given  angle. 

Solution.  Let  BAC  be  the  given 
angle. 

1.  From  the  vertex,  A,  of  the  angle 
BAC,  with  a  radius,  AE,  taken  at 
pleasure,  describe  an  arc  of  a  circle  ; 
and  from  the  two  points  D  and  E, 
where  this  arc  cuts  the  legs  of  the 
given  angle,  with  the  same  radius 
describe  two  other  arcs,  cutting  each 
other  in  the  point  m. 


144  GEOMETRY. 

2.  Through  the  point  m,  and  the  vertex  of  the  given 
angle,  draw  a  straight  line,  Am,  which  will  bisect  the 
given  angle  BAC. 

Demon.  The  two  triangles  AmD,  AmE,  have  the  three  sides 
in  the  one  equal  to  the  three  sides  in  the  other,  viz. 

the  side  AD  =  to  the  side  AE 

«     «     ?»D="    "     "     mE 

"     "      Am="    "      "     Am; 
consequently  these  two  triangles  are  equal  to  each  other ;  and  the 
angle  x,  opposite  to  the  side  mD,  in  the  triangle  AmD,  is  equal  to 
the  angle  y,  opposite  to  the  equal  side  mE,  in  the  triangle  AmE; 
therefore  the  angle  BAC  is  bisected. 

Problem  IV.  From  a  given  point  without  a  straight 
line,  to  let  fail  a  perpendicular  upon  that  line. 

Solution.  Let  A  be  the 
given  point,  from  which 
a  perpendicular  is  to  be 
drawn  to  the  line  MN. 

1.  With  any  radius  suf- 
ficiently great  describe  an 
arc  of  a  circle. 

2.  From  the  two  points  B  and  C,  where  this  arc  cuts 
the  line  MN,  draw  the  straight  lines  BA,  CA. 

3.  Bisect  the  angle  BAC  (see  the  last  Problem),  the 
line  AD  is  perpendicular  to  the  line  MN. 

Demon.  The  two  triangles  ABD,  ACD,  have  two  sides,  AB, 
AD,  in  the  one,  equal  to  two  sides,  AC,  AD,  in  the  other,  each  to 
each  (AC,  AB,  being  radii  of  the  same  circle,  and  the  side  AD 
being  common  to  both)  ;  and  have  the  angles  included  by  these 
sides  also  equal  (because  the  angle  BAC  is  bisected);  therefore 
these  two  triangles  are  equal  to  one  another  (Query  1,  Sect.  II.)  ; 
and  the  angle  y,  opposite  to  the  side  AB,  in  the  triangle  ABD,  is 
equal  to  the  angle  x,  opposite  to  the  equal  side  AC,  in  the  triangle 
ACD.    Now,  as  the  two  adjacent  angles  x  and  y,  which  the  straight 


..* 


GEOMETRY. 


145 


line  AD  makes  with  the  straight  line  MN,  are  equal  to  each  other, 
the  line  AD  must  be  perpendicular  to  MN.  (Def.  of  perpendicu- 
lar lines.) 

Problem  V.      To  bisect  a  given  straight  line. 

Solution.  Let  AB  be  the 
given  straight  line. 

1*  From  A,  with  a  radius 
greater  than  half  of  AB,  de- 
scribe an  arc  of  a  circle ;  and 
from  B}  with  the  same  radius, 
another,  cutting  the  first  in 
the  point  C. 

2.  From  the  point  C  draw  ' 
the  perpendicular  CM,  and  the  line  AB  is  bisected  in  M. 

Demon.  The  two  right-angled  triangles  AMC,  BMC,  are 
equal,  because  the  hypothenuse  AC  and  the  side  CM  in  the  one, 
are  equal  to  the  hypothenuse  and  the  side  CM  in  the  other 
(page  47) ;  and  therefore  the  third  side  AM  in  the  one,  is  also 
equal  to  the  third  side  BM  in  the  other;  consequently  the  line 
AB  is  bisected  in  the  point  M. 

Problem  VI.      To  transfer  a  given  angle. 

Solution.  Let  x  be  the 
given  angle,  and  A  the 
point  to  which  it  is  to  be 
transferred. 

1.  From  the  vertex  of  the 
given   angle,  as  a  centre, 

with  a  radius  taken  at  pleasure,  describe  an  arc  of  a  cir- 
cle between  the  legs  AB,  AC. 

2.  From  the  point  a,  as  a  centre,  with  the  same  radius, 
describe  another  arc,  cb. 

3.  Upon  trie  last  arc  take  a  distance,  6c,  equal  to  the 
chord  BC. 


146  GEOMETRY. 

4.  Through  a  and  c  draw  a  straight  line  ;  the  angle  y 
is  equal  to  the  angle  x. 

Demon.  The  arcs  BC,  6c,  are,  by  construction,  equal  to  one 
another ;  therefore  the  angles  x  and  y,  at  the  centre,  being  meas- 
ured by  these  arcs,  are  also  equal  to  one  another  (page  106, 1st). 

Problem  VII.  Through  a  given  point  draw  a  line 
parallel  to  a  given  straight  line. 


'y 


jp- 


Solution.  Let  E  be  the  point,  through  which  a  line 
is  to  be  drawn  parallel  to  the  straight  line  AB. 

1.  Take  any  point,  F,  in  the  straight  line  AB,  and  join 
/     EF. 

2.  *In  E  make  the  angle  y  equal  to  the  angle  % ;  the 
line  EG,  extended,  is  parallel  to  the  line  AB. 

Demon.  The  two  straight  lines  CG,  AB,  are  cut  by  a  third 
line  EF,  so  as  to  make  the  alternate  angles  x  and  y  equal ;  there- 
fore these  two  lines  are  parallel  to  each  other  (page  29, 2dly). 


Mechanical  Solution.  Take  a  ruler,  MN,  and  put 
it  in  such  a  position  that  a  right-angled  triangle,  passing 
along  its  edge,  as  you  see  in  the  figure,  will  make  with 
it,  in  different  points,  A,  C,  &c,  the  lines  AB,  CD,  &c. 


GEOMETRY. 


147 


These  lines  are  parallel  to  each  other,  because  they  are 
cut  by  the  edge  of  the  ruler  at  equal  angles.* 


Problem  VIII.  Two  adjacent  sides  and  the  angle 
included  by  them  being  given,  to  construct  a  parallelo- 
gram. 

A 


c 

A— 

1> 

/ 

B 

Solution.  Let  AB  and  AC  be  the  two  sides  of  true 
parallelogram,  and  x  the  angle  included  by  them. 

1.  Make  an  angle  equal  to  x. 

2.  Make  the  leg  AB  of  that  angle  equal  to  AB,  and 
the  leg  AC  equal  to  AC. 

3.  Through  the  point  C  draw  CD  parallel  to  AB,  and 
through  B,  the  line  BD  parallel  to  AC;  the  quadrilateral 
ABCD  is  the  required  parallelogram. 

Demon.  The  opposite  sides  of  the  quadrilateral  ABCD,  are 
parallel  to  each  other ;  therefore  the  figure  is  a  parallelogram. 
(See  Def.  page  13.) 


*  This  is  a  hetter  way  of  drawing  parallel  lines  than  the  common 
method  by  a  parallel  ruler,  which  is  seldom  very  accurate,  on 
account  of  the  instrument  being  frequently  out  of  order,  and  the 
great  steadiness,  of  hand  required  in  the  use  of  it 


148 


GEOMETRY. 


Problem  IX.     To  divide  a  given  line  into  any  number 
of  equal  parts. 

I.  Solution.  Let  AB  be  the 
given  line,  and  let  it  be  required  to 
divide  it  into  five  equal  parts. 

1.  From  the  point  A,  draw  an 
indefinite  straight  line  AN,  making 
any  angle  you  please  with  the  line 
AB. 

2.  Take  any  distance  Am,  and 
measure  it  off  5  times  upon  the  line 
AN. 

3.  Join  the  last  point  of  division  q, 
and  the  extremity  B  of  the  line  AB. 

4.  Through  m,  n,  o,  p,  q,  draw  the  straight  limes  bm, 
en,  do,  ep,  parallel  to  Bq;  the  line  AB  is  divided  into  five 
equal  parts. 

The  demonstration  follows  immediately  from  Query  14,  Sect.  H. 

II.  Solution.     Let  AB  be  tfte  — 

given  straight  line,  which  is  to  be 
divided  into  5  equal  parts. 

1.  Draw  a  straight  line  MN, 
greater  than  AB,  parallel  to  AB. 

2.  Take  any  distance  M?i,  and 
measure  it  off  5  times  upon  the 
line  MN. 

3.  Join  the  extremities  of  both 
the  lines  Mr  and  AB,  by  the  straight 
lines  MA,  rB,  which  will  cut  each 
other,  when  sufficiently  extended, 
in  a  point  I. 

4.  Join  In,  \o,  Ip,  \q,  the  line  AB  is  divided  into  6 
equal  parts,  viz.  A6,  6c,  cd,  de,  cB. 


GEOMETRY. 


149 


Demon.  The  triangles  A6I,  bcl,  cdl,  del,  eBI,  are  similar  to 
the  triangles  Mnl,  nol,  opl,pqI,  qrl,  each  to  each;  because  the 
line  AB  is  drawn  parallel  to  Mr  (Query  16,  Sect.  II.) ;  and  as  the 
bases  Mn,  no,  op,  pq,  qr,  of  the  latter  triangles  are  all  equal  to  one 
another,  the  bases  Ab,  be,  cd,  de,  eB  of  the  former  triangles  must 
also  be  equal  to  one  another. 

Remark.  If  it  were  required  to  divide  a  line  into  two  parts 
which  shall  be  in  a  given  ratio,  for  instance,  as  2  to  3,  you  need 
only,  as  before,  take  5  equal  distances  upon  the  line  MN,  and  then 
join  the  point  I  to  the  second  and  last  point  of  division ;  the  line 
AB  will,  in  the  point  c,  be  divided  in  the  ratio  of  2  to  3.  In  a  simi- 
lar manner  can  any  given  straight  line  be  divided  into  3,  4,  *>,  &c. 
parts,  which  shall  be  to  each  other  in  a  given  ratio. 


Problem  X.      Three  lines  being  given,  to  find  a  fourth 
one,  which  shall  be  in  a  geometrical  proportion  with  them. 


AAA 


D 


Solution.  Let  AB,  AC, 
AD,  be  the  given  straight 
lines,  which  are  three  terms 
of  a  geometrical  proportion, 
to  which  the  fourth  term  is 
wanting.  (See  Theory  of 
Proportions,  Principle  8th, 
page  64.) 

1.  Draw  two  indefinite 
straight  lines  AP,  AQ,,  mak- 
ing with  one  another  any  angle  you  please. 

2.  Upon  one  of  these  lines  measure  off  the  two  dis- 
tances AB,  AC,  and  on  the  other  the  distance  AD. 

3.  Join  BD,  and  through  C  draw  CE  parallel  to  BD ; 
the  line  AE  is  the  fourth  term  in  the"  geometrical  pro- 
portion 

AB:  AC  =  AD:  AE. 

Demon.    The  triangle  ABD  is  similar  to  the  triangle  ACE, 
from  which  it  may  be  considered  as  cut  off  by  the  line  BD  being 
13* 


150  GEOMETRY. 

drawn  parallel  to  CE  (Query  16,  Sect.  II.) ;  and  as  in  similar  tri- 
angles the  corresponding  sides  are  in  a  geometrical  proportion 
(page  70,  4thly),  we  have 

AB:  AC  =  AD:  AE. 


Problem  XI.  Two  angles  of  a  triangle  being  given, 
to  find  the  third  one. 

Solution.  Let  x  and  y  be 
the  two  given  angles  of  the 
triangle,  and  let  it  be  requir- 
ed to  find  the  third  angle  z. 

In  any  point  O  of  an  in- 
definite   straight    line   AB, 
make  two  angles  x  and  y, 
equal  to  the  two  given  angles  of  the  triangle ;  the  remain- 
ing angle  z  is  equal  to  the  angle  z  in  the  triangle. 

Demon.  The  sum  of  the-  three  angles  x,  z,  y,  in  the  triangle, 
is  equal  to  two  right  angles  (Query  13,  Sect.  I.),  and  the  sum  of 
the  three  angles  x,  y,  z,  made  in  the  same  point  O,  and  on  the 
same  side  of  the  straight  line  AB,  is  also  equal  to  two  right  angles 
(page  23)*;  and  as  the  angles  x  and  y  are  made  equal  to  the  angle 
x  and  y  in  the  triangle,  the  remaining  angle  z  is  also  equal  to  the 
remaining  angle  z  in  the  triangle. 

Remark.  If,  instead  of  the  angles  themselves,  their  measure 
were  given  in  degrees,  minutes,  seconds,  &c,  you  need  only  sub- 
tract the  sum  of  the  two  angles  from  180  degrees,  which  is  the 
measure  of  two  right  angles  ;  the  remainder  is  the  angle  sought 

Problem  XII.  Through  three  given  points,  which  are 
not  in  the  same  straight  line,  to  describe  the  circumference 
of  a  circle. 

Solution.  Let  A,  B,  C,  be  the  three  points,  through 
which  it  is  required  to  pass  the  circumference  of  a  circle. 

1.  Join  the  three  points  A,  B,  C,  by  the  straight  lines 
AB,  BC. 


GEOMETRY.  151 

2.  Bisect  the  lines  AB,  BC. 

3.  In  the  points  of  bisection  E 
and  F,  erect  the  perpendiculars 
EO,  FO,  which  will  cut  each  other 
in  a  point  O. 

4.  From  the  point  O  as  a  cen- 
tre, with  a  radius  equal  to  the  dis- 
tance AO,  describe  the  circumference  of  a  circle,  and  it 
will  pass  through  the  three  points  A>  B,  C. 

Demon.  The  two  points  A  and  B  are  at  an  equal  distance 
from  the  foot  of  the  'perpendicular  EO ;  therefore  AO  and  BO  are 
equal  to  one  another  (page  45,  5thly)  ;  for  the  same  reason  is  BO 
equal  to  OC  ;  because  the  points  B  and  C  are  at  an  equal  distance 
from  the  foot  of  the  perpendicular  FO ;  and  as  the  three  lines  AO, 
BO,  CO  are  equal  to  one  another,  the  three  points  A,  B,  C,  must 
necessarily  lie  in  the  circumference  of  the  circle  described  with 
the  radius  AO. 

Problem  XIII.  To  find  the  centre  of  a  circle,  or  of  a 
given  arc. 

Solution.  Let  the  circle  in  the  last  figure  be  the 
given  one. 

1.  Take  any  three  points  A,  B,  C,  in  the  circumfer- 
ence, and  join  them  by  the  chords  AB,  BC. 

2.  Bisect  each  of  these  chords,  and  in  the  points  of 
bisection  erect  the  perpendiculars  EO,  FO ;  the  point  O, 
in  which  these  perpendiculars  meet  each  other,  is  the 
centre  of  the  circle. 

In  precisely  the  same  manner  can  the  centre  of  an  arc 
be  found. 

The  demonstration  is  exactly  the  same  as  in  the  last  problem. 


152 


GEOMETRY. 


Problem  XIV.    In  a  given  point  in  the  circumference 
of  a  circle,  to  draw  a  tangent  to  that  circle. 

Solution.  Let  A  be  the  given 
point  in  the  circumference  of  the 
circle. 

Draw  the  radius  AO,  and  at  the 
extremity  A,  perpendicular  to  it,  the 
line  MN  ;  and  it  is  a  tangent  to  the 
given  circle. 

Demon.  The  line  MN  being  drawn  at 
the  extremity  AO  of  the  radius,  and  per- 
pendicular to  it,  touches  the  circumference 
in  only  one  point  (page  108) 


Problem  XV.  From  a  given  point  without  a  circle, 
to  draw  a  tangent  to  the  circle. 

Solution.  Let  A  be  the  given 
point,  from  which  a  tangent  is  to 
be  drawn  to  the  circle. 

1.  Join  the  point  A  and*  the 
centre  C,  of  the  given  circle. 

2.  From  the  middle  of  the  line 
AC  as  a  centre,  with  a  radius  equal 
to  BC  =  AB,  describe  the  circum- 
ference of  a  circle. 

3.  Through  the  points  E  and  D, 
where  this  circumference  cuts  the 
circumference  of  the  given  circle,  draw  the  lines  AD, 
AE  ;  and  they  are  tangents  to  the  given  circle. 

Demon.  Join  DC,  EC.  The  angles  x  and  y,  being  both  an- 
gles at  the  circumference  of  the  circle  whose  centre  is  B,  measure 
each  half  as  many  degrees  as  the  arc  on  which  their  legs  stand. 
Both  the  angles,  x  and  y,  have  their  legs  standing  on  the  diameter 
AC,  of  the  circle  B ;  therefore  each  of  these  angles  measures  half 


GEOMETRY. 


1S3 


&s  many  degrees  as  the  semi-circumference  (page  107,  Rem.  3d)  ; 
consequently,  they  are  both  right  angles,  and  the  lines  AE  and 
DA,  being  perpendicular  to  the  radii  CE,  DC,  are  both  tangents 
to  the  circle  C 

Remark.    From  a  point  without  a  circle,  you  can  always  draw 
two  tangents  to  the  same  circle. 


Problem  XVI.      To  draw  a  tangent  common  to  two 
given  circles. 

Solution.  Let  A  and  B  be  the 
centres  of  the  given  circles,  and  let 
it  be  required  to  draw  a  tangent, 
which  shall  touch  the  two  circles  on 
the  same  side. 

1.  Join  the  centres  of  the  two 
given  circles  by  the  straight  line 
AB. 

2.  From  B,  as  a  centre,  with  a 
radius  equal  to  the  difference  be- 
tween the  radii  of  the  given  circles, 
describe  a  third  circle. 

3.  From  A  draw  a  tangent  AE 
to  that  circle  (see  the  last  problem). 

4.  Draw  the  radius  BE,  and  extend  it  to  D. 

5.  Draw  the  radius  AC  parallel  to  BD. 

6.  Through  C  and  D  draw  a  straight  line,  and 
be  a  tangent  common  to  the  two  given  circles. 


it  will 


Demon.  The  radius  AC  being  equal  and  parallel  to  ED,  it 
follows  that  ACED  is  a  parallelogram ;  and  because  the  tangent 
AE  is  perpendicular  to  the  radius  BE  (page  108,  1st),  CD  is  per- 
pendicular to  BD ;  consequently  also  to  AC  (because  AC  is  parallel 
to  BD) ;  and  the  line  CD,  being  perpendicular  to  both  the  radii 
AC,  BD,  is  a  tangent  common  to  the  two  given  circles. 


154 


GEOMETRY. 


If  it  be  required  to  draw  a  tangent 
common  to  two  given  circles,  which 
shall  touch  them  on  opposite  sides,  then 

1.  From  B  as  a  centre,  with  a  radius 
equal  to  the  sum  of  the  radii  of  the  given 
circles,  describe  a  third  circle. 

2.  From  A  draw  a  tangent  AE  to  that 
circle. 

3.  Join  BE,  cutting  the  given  circle 
in  D. 

4.  Draw  AC  parallel  to  BE. 

5.  Through  C  and  D,  draw  a  straight 
line,  and  it  is  the  required  tangent,  touching  the  circles  on  opposite 
sides. 

The  demonstration  is  the  same  as  the  last. 


Problem  XVII.  Upon  a  given  straight  line  to  describe 
a  segment  of  a  circle,  ichich  shall  contain  a  given  angle  ; 
that  is,  a  segment,  such  that  the  inscribed  angles,  having 
their  vertices  in  the  arc  of  the  segment  and  their  legs 
standing  on  its  extremities,  shall  each  be  equal  to  a  given 


Solution.  Let  AB  be  the 
given  line,  and  x  the  given 
angle. 

1.  Extend  AB  towards  C. 

2.  Transfer  the  angle  x  to 
the  point  A. 

3.  Bisect  AB  in  E. 

4.  From  the  points  A  and  &  "£" 
E,  draw  the  lines  AO  and  EO, 

respectively,  perpendicular  to  FG  and  CB. 

5.  From  the  point  O,  the  intersection  of  these  perpen- 
diculars, as  a  centre,  with  a  radius  equal  to  OA,  describe 
a  circle ;  AMNB  is  the  required  segment. 


GEOMETRY. 


155 


Demon;  The  line  FG  being,  by  construction,  perpendicular  to 
the  radius  AO,  is  a  tangent  to  the  circle  (page  108) ;  and  the  angle 
GAB,  formed  by  that  tangent  and  the  chord  AB,  is  equal  to  either 
of  the  angles  AMB,  ANB,  &c,  that  can  be  inscribed  in  the  seg- 
ment AMNB ;  because  the  angle  GAB  measures  half  as  many  de- 
grees as  the  arc  ALB  (page  109),  and  each  of  the  angles  AMB, 
ANB,  &c,  at  the  circumference,  having  its  legs  standing  on  the 
extremities  of  the  chord  AB,  measures  also  half  as  many  degrees  as 
the  arc  ALB  (page  111) ;  and  as  the  angle  GAB  is  equal  to  the  angle 
x,  GAB  and  x  being  opposite  angles  at  the  vertex  (Query  5, 
Sect.  I.),  each  of  the  angles  AMB,  ANB,  &c,  is  also  equal  to  the 
given  angle  x. 

Remark.  If  the  angle  a:  is  a  right 
angle,  the  segment  AMB  is  a  semi- 
circle, and  the  chord  AB  a  diameter. 
To  finish  the  construction,  you  need 
only  from  the  middle  of  the  line  AB  as 
a  centre,  with  a  radius  equal  to  OA, 
describe  a  semicircle,  and  it  is  the  re- 
quired segment ;  for  the  angle  AMB 

at  the  circumference  measuring  half  as  many  degrees  as  the  semi- 
circumference  AB,  on  which  its  legs  stand,  is  a  right  angle. 


D 


A 


BM 


a 


Problem  XVIII.  To  find  a  mean  proportional  (see 
page  66)  to  two  given  straight  lines. 

Solution.  LetAB,BC,       * 
be  the  two  given  lines. 

1.  Upon    an    indefinite 
straight  line,  take  the  two      B 
distances  AB,  BC. 

2.  Bisect  the  whole  dis- 
tance AC,  and  from  M,  the  middle  of  AC,  with  a  radius 
equal  to  AM,  describe  a  semi-circumference. 

3.  In  B  erect  a  perpendicular  to  the  diameter  AC,  and 
extend  it  until  it  meets  the  semi-circumference  in  D ;  the 
line  DB  is  a  mean  proportional  between  the  lines  AB 
and  BC. 


156  GEOMETRY. 

Demon.  The  triangle  ADC  is  right-angled  in  D ;  because  the 
angle  ADC  is  inscribed  in  a  semicircle  (see  the  remark  to  the  last 
problem) ;  and  the  perpendicular  DB  let  fall  from  the  vertex  D,  of 
the  right  angle  upon  the  hypothenuse,  is  a  mean  proportional  be- 
tween the  two  parts  AB,  BC,into  which  it  divides  the  hypothenuse 
(page  75,  1st)  ;  therefore  we  have  the  proportion 
AB:BD  =  BD:BC. 

Problem  XIX.  To  divide  a  given  straight  line  into 
two  such  parts,  that  the  greater  of  them  shall  be  a  mean 
proportional  between  the  smaller  part  and  the  whole  of  the 
given  line. 

Solution.  Let  AB  be  the 
given  straight  line. 

1.  At  the  extremity  B  of  the 
given  line,  erect  a  perpendicu- 
lar, and  make  it  equal  to  half  of 
the  line  AB. 

2.  From  O,  as  a  centre,  with  a  radius  equal  to  OB, 
describe  a  circle. 

3.  Join  the  centre  O  of  that  circle,  and  the  extremity 
A  of  the  given  line,  by  the  straight  line  AO. 

4.  From  AB  cut  off  a  distance  AD  equal  to  AE  ;  then 
AD  is  a  mean  proportional  between  the  remaining  part 
BD,  and  the  whole  line  AB ;  that  is,  you  have  the  pro- 
portion 

AB  :  AD  =  AD:BD. 

Demon-.  Extend  the  line  AO  until  it  meets  the  circumference 
in  C.  Then  the  radius  OB,  being  perpendicular  to  the  line  AB, 
we  have  from  the  same  point  A,  a  tangent  AB,  and  another  line 
AC  drawn  cutting  the  circle ;  therefore  we  have  the  proportion 

AC  :  AB  =  AB  :  AE ; 
for  the  tangent  AB  is  a  mean  proportional  between  the  whole  line 
AC,  and  the  part  AE  without  the  circle.  (Query  13,  Sect.  IV.) 

Now,  in  every  geometrical  proportion,  you  can  add  or  subtract 
the  second  term  once  or  any  number  of  times  from  the  first  term, 


GEOMETRY.  157 

and  the  fourth  term  the  same  number  of  times  from  the  third  term, 
•without  destroying  the  proportion  (page  62,  6th).  According  to 
this  principle  you  have 

AC  — AB  :  AB  =  AB  —  AE  :  AE; 
that  is,  the  line  AC  less  the  line  AB,  is  to  the  line  AB,  as  the  line 
AB  less  EA,  is  to  the  line  AE.     But  AC  less  AB  is  the  same  as 
the  line  AC  less  the  diameter  CE  (because  the  radius  of  the  circle 
is,  by  construction,  equal  to  half  the  line  AB) ;  and  AB  less  AE, 
is  the  same  as  AB  less  AD  (because  AD  is  made  equal  to  AE); 
therefore  you  may  write  the  above  proportion  also 
AE  :  AB  =  BD  :  AE,*  or  also 
AD  :  AB  =  BD  :  AD; 
and  because  in  every  geometrical  proportion  the  order  of  the  terms 
may  be  changed  in  both  ratios  (Principle  1,  of  Geom.  Prop.),  you 
can  change  the  last  proportion  into 

AB  :  AD  =  AD  :  BD; 
that  is,  the  part  AD  of  the  line  AB,  is  a  mean  proportional  between 
the  whole  line  AB  and  the  remaining  part  BD. 

Problem  XX.    To  inscribe  a  circle  in  a  given  triangle. 
Solution.     Let   the   given 
triangle  be  ABC. 

1.  Bisect  two  of  the  angles  "2//CT^N?1' 
of  the  given  triangle;  for  in- 
stance the  angles  at  C  and  B,       a* 
by  the  lines  CO,  BO. 

2.  From  the  point  O,  where  these  lines  cut  each  other, 
let  fall  a  perpendicular  upon  any  of  the  sides  of  the  given 
triangle. 

3.  From  O,  as  a  centre,  with  the  radius  OP,  equal  to 
the  length  of  that  perpendicular,  describe  a  circle,  and  it 
will  be  inscribed  in  the  triangle  ABC. 

Demon.  From  0  let  fall  the  perpendiculars  OM,  ON,  upon 
the  two  sides  BC,  AC,  of  the  given  triangle.     The  angle  OCM  is, 

*  AC  less  the  diameter  CE,  being  equal  to  AE  ;  and  BA  less  AD, 
equal  to  BD. 

14 


158  GEOMETRY. 

by  construction,  equal  to  the  angle  OCN  (because  the  angle  ACB 
is  bisected  by  the  line  CO)  ;  and  CMO,  CNO,  being  right  angles, 
the  angles  COM  and  CON  are  also  equal  to  one  another  (because 
when  two  angles  in  one  triangle  are  equal  to  two  angles  in  another, 
the  third  angles  in  these  triangles  are  also  equal) ;  therefore  the 
two  triangles  CMO,  CNO,  have  a  side  CO,  and  the  two  adjacent 
angles  in  the  one,  equal  to  the  same  side  CO,  and  the  two  adjacent 
angles  in  the  other ;  consequently  these  two  triangles  are  equal  to 
one  another ;  and  the  side  OM,  opposite  to  the  angle  OCM  in  the 
one,  is  equal  to  the  side  ON,  opposite  to  the  equal  angle  OCN  in 
the  other.  In  the  same  manner  it  may  be  proved  that  the  perpen- 
dicular OM  is  also  equal  to  OP ;  and  as  the  three  perpendiculars 
OM,  ON,  OP,  are  equal  to  one  another,  the  circumference  of  a 
circle  described  from  the  point  0  as  a  centre,  with  a  radius  equal 
to  OP,  passes  through  the  three  points  M,  N,  P ;  and  the  sides 
AB,  BC,  AC  of  the  given  triangle,  being  perpendicular  to  the  radii 
OP,  OM,  ON,  are  tangents  to  the  inscribed  circle  (page  108). 

Problem  XXI.  To  circumscribe  a  circle  about  a  tri~ 
angle. 

This  problem  is  the  same  as  to  make  the  circumfer- 
ence of  a  circle  pass  through  three  given  points.  (See 
Problem  XII.) 

Problem  XXII.      To  trisect  a  right  angle. 

Solution.  Let  BAC  be  the 
right  angle  which  is  to  be  divid- 
ed into  three  equal  parts. 

1.  Upon  AB  take  any  dis- 
tance AD,  and  construct  upon 
it  the  equilateral  triangle  ADE. 
(Problem  I.) 

2.  Bisect  the  angle  DAE  by  the  line  AM  (Problem 
III.)  ;  and  the  right  angle  BAC  is  divided  into  the  three 
equal  angles  CAE,  EAM,  MAB. 


GEOMETRY. 


159 


Demon-.  The  angle  BAE  being  one  of  the  angles  of  an  equi- 
lateral triangle,  is  one  third  of  two  right  angles  (page  33),  and 
therefore  two  thirds  of  one  right  angle  ;  consequently  CAE  is  one 
third  of  the  right  angle  BAC ;  and  since  the  angle  BAE  is  bisected 
by  the  line  AM,  the  angles  EAM,  MAB,  are  each  of  them  also 
equal  to  one  third  of  a  right  angle  ;  and  are  therefore  equal  to  the 
angle  CAE  and  to  each  other. 


PART  II. 


Problems  relative  to  the  transformations  of  geomet- 
rical figures. 

Problem  XXIII.  To  transform  a  giv-en  quadrilateral 
figure  into  a  triangle  of  equal  area,  whose  vertex  shall 
be  in  a  given  angle  of  the  figure,  and  whose  base  in  one 
of  the  sides  of  the  figure. 


B 

Fig.  I. 

^E 

Fig.  IL 
B 

A 

I) 

A 


JE 


Solution.  Let  ABCD  (Fig.  1.  and  II.),  be  the  given 
quadrilateral ;  the  figure  I.  has  all  its  angles  outwards, 
and  the  figure  II.  has  one  angle,  BCD,  inwards  ;  let  the 
vertex  of  the  triangle,  which  shall  be  equal  to  it,  fall  in  B, 

1.  Draw  the  diagonal  BD  (Fig.  I.  and  II. ),  and  from 
C,  parallel  to  it,  the  line  CE. 

2.  From  E,  where  the  line  CE  cuts  AD  (Fig.  II.),  or 
its  further  extension  (Fig.  I.),  draw  the  line  EB;  the 
triangle  ABE  is  equal  to  the  quadrilateral  ABCD. 


160 


GEOMETRY. 


Demon.  The  area  of  the  triangle  BCD  (Fig.  I.  and  II.)  w 
equal  to  the  area  of  the  triangle  BDE ;  because  these  two  triangles 
are  upon  the  same  basis,  BD,  and  between  the  same  parallels,  BD, 
CE  (page  90,  3dly) ;  consequently  (Fig.  I.),  the  sum  of  the  areas 
of  the  two  triangles  ABD  and  BDC,  is  equal  to  the  sum  of  the 
areas  of  the  two  triangles  ABD,  BDE ;  that  is,  the  area  of  the 
quadrilateral  ABCD  is  equal  to  the  sum  of  the  areas  of  the  two  tri- 
angles ABD,  BDE,  which  is  the  area  of  the  triangle  ABE. 

And  in  figure  II.  the  difference  between  the  areas  of  the  two 
triangles  ABD,  BCD,  that  is,  the  quadrilateral  ABCD,  is  equal 
to  the  difference  between  the  triangles  ABD,  EBD,  which  is  the 
triangle  ABE. 

Problem  XXIV.  To  transform  a  given  pentagon  into 
a  triangle,  whose  vertex  shall  be  in  a  given  angle  of  the 
pentagon,  and  whose  base  upon  one  of  its  sides. 

Solution.  Let  ABCDE  (Fig.  I.  and  II.),  be  the  given 
pentagon ;  let  the  vertex  of  the  triangle,  which  is  to  be 
equal  to  it,  be  in  C. 

Fig.  I.  Fig.  II. 


1.  From  C  draw  the  diagonals  CA,  CE. 

2.  From  B  draw  BF  parallel  to  CA,  and  from  D  draw 
DG  parallel  to  CE. 

3.  From  F  and  G,  where  these  parallels  cut  AE  or  its 
further  extension,  draw  the  lines  CF,  CG ;  CFG  is  the 
triangle  required. 

Demon.  In  both  figures,  we  have  the  area  of  the  triangle 
CBA  equal  to  the  area  of  the  triangle  CFA  ;  because  these  two 
triangles  are  upon  the  same  basis,  CA,  and  between  the  same  par- 


GEOMETRY. 


161 


allels,  AC,  FB ;  and  for  the  same  reason  is  the  area  of  the  triangle 
CDE  equal  to  the  area  of  the  triangle  CGE ;  therefore  in  figure  I. 
the  sum  of  the  areas  of  the  three  triangles  CAE,  CBA,  CDE,  is 
equal  to  the  sum  of  the  areas  of  the  triangles  CAE,  CFA,  CGE ;  that 
is,  the  area  of  the  pentagon  ABCDE  is  equal  to  the  area  of  the 
triangle  CFG;  and  in  figure  II.  the  difference  between  the  area 
of  the  triangle  CAE  and  the  areas  of  the  two  triangles  CBA,  CDE, 
is  equal  to  the  difference  between  the  area  of  the  same  triangle 
CAE*  and  the  areas  of  the  two  triangles  CFA,  CGE ;  that  is,  the 
area  of  the  pentagon  ABCDE  is  equal  to  the  area  of  the  triangle 
CFG. 


Problem  XXV.  To  convert  any  given  figure  into  a 
triangle,  whose  vertex  shall  be  in  a  given  angle  of  the 
figure,  and  whose  basis  shall  fall  upon  one  of  its  sides. 


Fig.  II. 


Let  ABCDEF  (Fig.  I.  and  II.)  be  the  given  figure 
(in  this  case  a  hexagon),  and  A  the  angle  in  which  the 
vertex  of  the  required  triangle  shall  be  situated.  For  the 
sake  of  perspicuity,  I  shall  enumerate  the  angles  and  sides 
of  the  figure  from  A,  and  call  the  first  angle  A,  the  sec- 
ond B,  the  third  C,  and  so  on  ;  further,  AB  the  first  side, 
BC  the  second,  DE  the  third,  and  so  on.  We  shall  then 
have  the  following  general  solution. 

1.  From  A  to  all  the  angles  of  the  figure,  draw  the 
diagonals  AC,  AD,  AE,  which,  according  to  the  order 
in  which  they  stand  here,  call  the  first,  second,  and  third 
diagonal. 

14* 


162  GEOMETRY. 

2.  Draw  from  the  second  angle,  B,  a  line,  Ba,  parallel 
to  the  first  diagonal,  AC ;  from  the  point  where  the  par- 
allel meets  the  third  side,  CD  (Fig.  II.),  or  its  further 
extension  (Fig.  I.),  draw  a  line,  ab,  parallel  to  the  second 
diagonal,  AD ;  and  from  the  point  b,  where  this  meets  the 
fourth  side  DE  (Fig.  II.)  or  its  further  extension  (Fig.  I.), 
draw  another  line,  be,  parallel  to  the  third  diagonal. 

3.  When,  in  this  way,  you  have  drawn  a  parallel  to 
every  diagonal,  then,  from  the  last  point  of  section  of  the 
parallels  and  sides  (in  this  case  c),  draw  the  line  cA ; 
AcF  is  the  required  triangle,  whose  vertex  is  in  A,  and 
whose  basis  is  in  the  side  EF. 

The  demonstration  is  similar  to  the  one  given  in  the  two  last 
problems.  First,  each  of  the  hexagons  is  converted  into  the  pen- 
tagon AaDEF ;  then  the  pentagon  AaDEF  into  a  quadrilateral, 
A6EF  ;  and  finally  this  quadrilateral  into  the  triangle  AcF.  The 
areas  of  these  figures  are  evidently  equal  to  one  another  ;  for  the 
areas  of  the  triangles,  which,  by  the  above  construction,  are  suc- 
cessively cut  off,  are  equal  to  the  areas  of  the  new  triangles  which 
are  successively  added  on.  (See  the  demonstration  of  the  last 
problem.) 

Remark.  Although  the  solution  given  here  is  only  intended  for 
a  hexagon,  yet  it  may  easily  be  applied  to  every  other  rectilinear 
figure.  All  depends  upon  the  substitution  of  one  triangle  for 
another,  by  means  of  parallel  lines.  It  is  not  absolutely  necessary 
actually  to  draw  the  parallels ;  it  is  Only  requisite  to  denote  the 
points  in  which  they  cut  the  sides,  or  their  further  extension;  be- 
cause all  depends  upon  the  determination  of  these  points. 

Problem  XXVI.  To  transform  any  given  figure  into 
a  triangle  ivhose  vertex  shall  be  in  a  certain  point,  in  one 
of  the  sides  of  the  figure,  or  within  it,  and  whose  base 
shall  fall  upon  a  given  side  of  the  figure. 


GEOMETRY, 


Solution.  1st  Case.  Let  ABCDEF  be  a  hexagon, 
which  is  to  be  transformed  into  a  triangle  ;  let  the  vertex 
of  the  triangle  be  in  the  point  M  in  the  side  CD,  and  the 
base  in  AF. 

1.  In  the  first  place,  get  rid  of  the  angle  ABC,  by 
drawing  Ba  parallel  to  CA,  and  joining  Ca;  the  triangle 
CBa,  substituted  for  its  equal  the  triangle  ABa  (for  these 
two  triangles  are  upon  the  same  basis,  aB,  and  between 
the  same  parallels,  CA,  Ba),  transforms  the  hexagon 
ABCDEF  into  the  pentagon  aCDEF. 

2.  Draw  the  lines  Ma,  MF,  and  the  pentagon  aCDEF 
is  divided  into  three  figures,  viz.  the  triangle  MaF,  the 
quadrilateral  MDEF  on  the  right,  and  the  triangle  MCa 
on  the  left. 

3.  Transform  the  quadrilateral  MDEF  and  the  triangle 
MCa  into  the  triangles  MoT,  M6a,  so  that  the  basis  may 
be  in  AF  (see  the  last  problem)  ;  the  triangle  Mbd  is 
equal  to  the  given  hexagon. 


164 


2d  Case.  Let  ABCDEF  be  the  given  figure ;  let  the 
vertex  of  the  required  triangle  be  situated  in  the  point  M 
within  the  figure,  and  let  the  base  fall  upon  AF. 

1.  From  M  to  any  angle  of  the  figure,  say  D,  draw  the 
line  MD,  and  draw  the  lines  MA,  MF,  by  which  means 
the  figure  ABCDEF  is  divided  into  the  triangle  MAF, 
and  the  figures  MDCBA ,  MDEF. 

2.  Then  transform  MDCBA  and  MDEF  into  the  tri- 
angles McA,  MeF,  whose  bases  are  in  the  continuation 
of  AF  5  the  triangle  cMe  is  equal  to  the  figure  ABCDEF. 

The  demonstration  follows  from  those  of  the  last  three  problems. 

Problem  XXVII.  To  transform  a  given  rectangle 
into  a  square  of  equal  area. 


A. 


E 

y                   ,„"'* 

D\\ 

B 


M      G 


Solution.     Let  ABCD  be  the  given  rectangle. 

1.  Extend  the  greater  side,  AB,  of  the  rectangle,  making 
BM  equal  to  BD. 

2.  Bisect  AM  in  O,  and,  from  the  point  O  as  a  centre, 
with  a  radius  AO,  equal  to  OM,  describe  a  semicircle. 


GEOMETRY.  165 

3.  Extend  the  side  BD  of  the  rectangle,  until  it  meets 
the  circle  in  E. 

4.  Upon  BE  construct  the  square  BEFG,  which  is  the 
square  sought. 

Demon.     The  perpendicular  BE  is  a  mean  proportional  be- 
tween AB  and  BM  (see  Problem  XVIII.)  ;  therefore  we  have  the 

proportion 

AB  :  BE  =  BE  :  BM  ; 

and  as,  in  every  geometrical  proportion,  the  product  of  the  means 

equals  that  of  the  extremes  (Theory  of  Prop.,  Principle  10,  page 

65),  we  have  the  product  of  the  side  BE  multiplied  by  itself,  equal 

to  the  product  of  the  side  AB  of  the  parallelogram,  multiplied  by 

the  adjacent  side  BD  (or  BM).     But  the  first  of  these  products 

is  the  area  of  the  square  BEFG,  and  the  other  is  the  area  of  the 

rectangle  ABCD.;  therefore  these  two  figures  are,  in  area,  equal 

to  one  another. 

Problem  XXVIII.      To  transform  a  given  triangle 
into  a  square  of  equal  area. 


K 


D    JB    M   JP 


Solution.  Let  ABC  be  the  given  triangle,  AB  its 
base,  and  CD  its  height. 

1.  Extend  AB  by  half  the  height  CD. 

2.  Upon  AM  as  a  diameter,  describe  a  semicircle. 

3.  From  B  draw  the  perpendicular  BN,  which  is  the 
side  of  the  square  sought. 

Demon.  From  the  demonstration  in  the  last  problem,  it  follows, 
that  the  square  upon  BN  is  equal  to  the  rectangle,  whose  base  is 
AB,  and  whose  height  is  BM  (half  the  height  of  the  triangle 
ABC).  But  the  triangle  ABC  is  equal  to  a  rectangle  upon  the 
same  base  AB,  and  of  half  the  height  CD  (page  89,  1st)  ;  therefor© 


166 


GEOMETRY. 


the  area  of  the  square  BNOP  is  equal  to  the  area  of  the  triangle 
ABC. 

Remark.  It  appears  from  this  problem,  that  every  rectilinear 
figure  can  be  converted  into  a  square  of  equal  area.  It  is  only 
necessary  to  convert  the  figure  into  a  triangle  (according  to  the 
rules  given  in  the  problems  23,  24,  25),  and  then  that  triangle  into 
a  square. 

Problem  XXIX.  To  convert  any  given  triangle  into 
an  isosceles  triangle  of  equal  area. 

Solution.  Let  ABC  be 
the  given  triangle,  which  is 
to  be  converted  into  an  isos- 
celes one. 

1.  Bisect  the  base  AC  in 
D,  and  from  D  draw  the  perpendicular  DE. 

2.  From  the  vertex,  B,  of  the  given  triangle,  draw  BE, 
parallel  to  the  base,  AC. 

3.  From  the  point  E,  where  this  parallel  meets  the 
perpendicular,  draw  the  straight  lines  EA,  EC  ;  EAC  is 
the  isosceles  triangle  sought. 

Demon.  The  triangles  AEC  and  ABC  are  upon  the  same  basis, 
AC,  and  between  the  same  parallels  (page  90,  3dly). 


Problem  XXX.  To  convert  a  given  isosceles  triangle 
into  an  equilateral  one  of  equal  area.  (This  problem  is 
intended  for  more  advanced  and  elder  pupils.) 

Solution.  Let  ABC  be  the 
given  isosceles  triangle. 

1.  Upon  the  base,  AC,  of  the 
given  triangle,  draw  the  equilate- 
ral triangle  AEC  (problem  T.)  ; 
and  through  the  vertices,  E,  B,  of 
the  two  triangles,  draw  the  straight 


GEOMETRY.  107 

line  EB,  which  evidently  is  perpendicular  to  AC,  and 
bisects  the  last  line  in  D  (ABC,  AEC,  being  isosceles 
triangles). 

2.  Upon  ED  describe  the  semicircle  EFD,  and  from 
B  draw  the  perpendicular  BF,  which  meets  the  semi- 
circle in  F. 

3.  From  D,  with  the  radius  DF,  describe  an  arc,  FG, 
cutting  the  line  DE  in  G. 

4.  From  G,  draw  the  lines  GH,  GI,  parallel  to  the 
sides  of  the  equilateral  triangle  AEC  ;  HGI  is  the  equi- 
lateral triangle  sought. 

Demon.  Since  the  line  GH  is  parallel  to  AE,  and  GI  parallel 
to  EC,  the  angle  GHI  is  equal  to  the  angle  EAI,  and  the  angle 
GIH  to  the  angle  ECH  (page  31).  Thus  the  two  triangles  GHI, 
AEC,  have  two  angles,  GHI,  GIH,  in  the  one,  equal  to  two  an- 
gles, EAC,  EGA,  in  the  other,  each  to  each  ;  consequently  they 
are  similar  to  each  other  (page  73,  1st)  ;  and  the  triangle  GHI 
must  also  be  equilateral. 

Suppose  the  lines  DF  and  EF  drawn ;  then  DF  is  a  mean  propor- 
tional between  DE  and  DB ;  for  the  triangle  EDF  is  right-angled 
(see  the  Remark,  page  155)  in  F,  and  if  from  the  vertex  of  the 
right  angle,  the  perpendicular  FB  is  let  fall  upon  the  hypothenuse, 
the  side  DF  is  a  mean  proportional  between  the  hypothenuse,  ED, 
and  the  part,  BD,  of  it,  between  the  foot  of  the  perpendicular,  and 
the  extremity,  D,  of  the  line  FD  (see  page  75,  2dly) ;  consequently 
we  shall  have  the  proportion 

ED  :  DF  =  DF  :  BD  ; 
and  as  DG  is,  by  construction,  made  equal  to  DF, 

ED  :  DG  =  DG  :  BD (I.) 

Moreover,  in  the  two  similar  triangles,  ADE,  HDG,  the  corre- 
sponding sides  are  proportional  (page  70,  4thly) ;  therefore  we  have 
the  proportion 

ED  :  DG=AD  :  HD (II.) 

This  last  proportion  has  the  first  ratio  common  with  the  first  pro- 
portion ;  consequently  the  two  remaining  ratios  are  in  a  geometri- 
cal proportion  (Theory  of  Prop.,  Prin.  3d)  ;  that  is,  we  have 

AD  :  HD  =  DG:  BD; 
and  as,  in  every  geometrical  proportion,  the  product  of  the  means  is 


168 


GEOMETRY. 


equal  to  that  of  the  extremes  (Theory  of  Prop.,  Principle  10th),  we 
have  HD  multiplied  by  DG,  equal  to  AD  multiplied  by  BD ;  con- 
sequently, also,  half  the  product  of  the  line  HD,  multiplied  by  the 
line  DG,  equal  to  half  the  product  of  the  line  AD,  multiplied  by 
BD.  But  half  the  product  of  the  line  HD,  multiplied  by  DG,  is 
the  area  of  the  triangle  HDG  ;  because  the  triangle  HDG  is  right- 
angled  in  D,  therefore  if  HD  is  taken  for  the  basis,  DG  is  its 
height ;  and  for  the  same  reason  is  half  the  product  of  the  line  AD 
by  BD,  the  area  of  the  triangle  ADB ;  consequently  the  areas  of 
the  two  triangles,  ADB  and  HDG,  are  equal  to  one  another;  and 
because  the  triangle  HDG  is  equal  to  the  triangle  IDG,  and  the 
triangle  ABD  to  the  triangle  CBD,  the  area  of  the  whole  triangle 
HIG  is  equal  to  the  area  of  the  whole  triangle  ABC  ;  therefore  the 
triangle  HIG  is  the  required  equilateral  triangle,  which  is  equal, 
in  area,  to  the  given  isosceles  triangle,  ABC. 

Remark  1.  If  BD  is  greater  than  ED,  then  the  perpendicular, 
BF,  does  not  meet  the  semicircle.  In  this  case,  it  is  necessary  to 
describe  the  semicircle  on  BD,  and  from  E  to  draw  the  perpen- 
dicular. In  this  case,  the  points  H,  I,  will  not  be  situated  in  the 
line  AC  ;  but  in  its  further  extension. 

Remark  2.  From  this  and  the  preceding  problems,  it  appears 
how  any  figure  may  be  converted  into  an  equilateral  triangle  ;  for 
it  is  only  necessary  first  to  convert  the  figure  into  a  triangle,  this 
triangle  into  an  isosceles  triangle,  and  the  isosceles  triangle  into 
an  equilateral  one. 


Problem  XXXI.     To  describe  a  square,  which  in  area 
shall  be  equal  to  the  sum  of  several  given  squares. 

E 

A  B  C  B 


JE 


D 


A  B     B 

Solution.  Let  AB,  BC,  CD,  DE,  be  the  sides  of  four 
squares;  it  is  required  to  find  a  square  which  shall  be 
equal  to  the  sum  of  these  four  squares. 


GEOMETRY.  169 

1.  At  the  extremity,  B,  of  the  line  AB,  draw  a  perpen- 
dicular equal  to  BC,  and  join  AC. 

2.  At  the  extremity,  C,  of  the  line  AC,  draw  a  perpen- 
dicular equal  to  CD,  and  join  AD. 

3.  At  the  extremity,  D,  of  the  line  AD,  draw  a  perpen- 
dicular equal  to  DE,  and  join  AE ;  the  square  upon  AE 
is,  in  area,  equal  to  the  sum  of  the  four  squares  upon  the 
lines  AB,  BC,  CD,  DE. 

Demon.  The  square  upon  the  hypothenuse,  AC,  of  the  right- 
angled  triangle  ABC,  is  equal  to  the  sum  of  the  squares  upon  the 
two  sides  AB,  BC  (Query  6,  Sect.  III.) ;  and  for  the  same  reason 
is  the  square  upon  AD  equal  to  the  sum  of  the  squares  upon  CD 
and  AC ;  consequently,  also,  to  the  squares  upon  CD,  CB,  and  AB 
(the  square  upon  AC  being  equal  to  the  squares  upon  CB  and  AB)  ; 
and  finally  the  square  upon  AE  is  equal  to  the  sum  of  the  squares 
upon  ED  and  AD  ;  or,  which  is  the  same,  to  the  sum  of  the  squares 
upon  DE,  CD,  CB,  and  AB 

Problem  XXXII.  To  describe  a  square  which  s?ialll>e 
equal  to  the  difference  of  two  given  squares. 

Solution.     Let  AB,  AC 
be  the  sides  of  two  squares. 

1.  Upon  the  greater  side, 

AB,  as  a  diameter,  describe 
a  semicircle. 

2.  From    A,  within   the 
semicircle,    draw    the    line 

AC,  equal  to  the  given  line  AC,  and  join  BC;  then  CB 
is  the  side  of  the  square  sought. 

Demon.     The  triangle  ABC  is  right-angled  in  C,  and  in  every 
right-angled  triangle,  the  square  upon  one  of  the  sides,  which 
include  the  right  angle,  is  equal  to  the   difference  between  the 
squares  upon  the  hypothenuse  and  the  other  side. 
15 


170 


GEOMETRY. 


Problem    XXXIII.       To  transform  a  given  figure 
in  such  a  way,  that  it  may  be  similar  to  another  figure. 


n   cL 


Solution.  Let  X  be  the  given  figure,  and  ABCDEF 
the  one  to  which  it  is  to  be  similar. 

1.  Convert  the  figure  ABCDEF  into  a  square  (seethe 
remark,  page  166),  and  let  its  side  bemw,  so  that  the  area 
of  the  square  upon  mn  is  equal  to  the  area  of  the  figure 
ABCDEF  ;  convert,  also,  the  figure  X  into  a  square,  and 
let  its  side  be  pq,  so  that  the  area  of  the  square  upon  pq 
shall  be  equal  to  the  area  of  the  figure  X. 

2.  Take  any  side  of  the  figure  ABCDEF,  say  AF ; 
and  to  the  three  lines,  mn,  pq,  AF,  find  a  fourth  propor- 
tional (Problem  X),  which  you  cut  off  from  AF.  Let 
Af  be  this  fourth  proportional,  so  that  we  have  the  pro- 
portion 

mn  :  pq  =  AF  :  Af. 

3.  Then  draw  the  diagonals  AE,  AD,  AC,  and  the  lines 
fe,  ed,  dc,  cb,  parallel  to  the  lines  FE,  ED,  DC,  CB  ; 
then  Abcdef  will  be  the  required  figure,  which  in  area  is 
equal  to  the  figure  X,  and  is  similar  to  the  figure  ABCDEF. 

Demon.  It  is  easily  proved,  that  the  figure  Abcdef  is  similar 
to  ABCDEF.  Further,  we  know  that  the  areas  of  the  two  similar 
figures,  ABCDEF,  Abcdef,  are  to  each  other,  as  the  areas  of  the 
squares  upon  the  corresponding  sides  AF,  A/,  (see  page  198); 
Which  may  be  expressed, 

ABCDEF  :  Abcdef  =  AF  X  AF  :  A/x  Af; 


GEOMETRY.  171 

and  as  the  sides  AF  and  A/  are  (by  construction  2)  in  proportion 
to  the  lines  mn,  pq,  the  squares  upon  these  sides,  and  therefore 
the  figures  ABCDEF,  Abcdef,  themselves,  are  in  proportion  to  the 
squares  upon  mn  and  pq  ;  that  is,  we  shall  have  the  proportion 
ABCDEF  :  Abcdef —mn  X  mn  : pq  X pq- 
This  proportion  expresses,  that  the  area  of  the  figure  ABCDEF 
is  as  many  times  greater  than  the  area  of  the  figure  Abcdef,  as  the 
area  of  the  square  upon  the  line  mn  is  greater  than  the  area  of 
the  square  upon  the  line  pq  ;  therefore,  as  the  area  of  the  figure 
ABCDEF  is,  by  construction,  equal  to  that  of  the  square  upon  the 
line  mn,  the  area  of  the  figure  Abcdef  is  equal  to  that  of  the 
square  upon  the  line  pq.  But  the  square  upon  pq  is  made  equal 
to  that  of  the  figure  X  ;  therefore  the  area  of  the  figure  Abcdef 
is  also  equal  to  that  of  the  figure  X  ;  and  the  figure  Abcdef  is  the 
one  required. 


PART  III. 

Partition  of  figures  by  drawing. 

Problem  XXXIV.  To  divide  a  triangle  from  one  of 
the  vertices  Jinto  a  given  number  of  parts. 

Solution.  Let  ABC  be  the  given  triangle,  which  is 
to  be  divided,  say,  into  six  equal  parts ;  let  A  be  the  ver- 
tex, from  which  the  lines  of  division  are  to  be  drawn. 

1.  Divide  the  side  BC,  . 
opposite  the  vertex  A,  into 
six  equal  parts,  BD,  DE,                     / 
EF,  FG,  GH,  HC.                           // 

2.  From  A  to  the  points  /  / 
of  division,  D,  E,  F,  G,  H,       BJDHL 
draw  the  lines  AD,  AE, 

AF,  AG,  AH;   the  triangle  ABC  is  divided  into  the  six 
equal  triangles,  ABD,  ADE,  AEF,  AFG,  AGH,  AHC. 


172 


GEOMETRY. 


Demon.    The  triangles  ABD,  ADE,  AEF,  AFG,  AGH,  AHC, 

are,  in  area,  equal  to  one  another,  because  they  have  equal  bases 
and  the  same  height,  Am  (page  89). 

Remark.  If  it  is  required  to  divide  the  triangle  ABC  according 
to  a  given  proportion,  it  will  only  be  necessary  to  divide  the  line 
BC  in  this  proportion,  and  from  A  to  draw  lines  to  the  points  of  di- 
vision. 


Problem  XXXV.  From  a  given  point  in  one  of  the 
sides  of  a  triangle,  to  divide  it  into  a  given  number  of 
equal  parts. 

Solution.  Let  ABC  be 
the  given  triangle,  which  is 
to  be  divided  into  eight  equal 
parts ;  the  lines  of  division 
are  to  be  drawn  from  T. 

1.  Make  Aa  and  Bb  equal 
to  J  of  AB,  and  from  T  draw   . 
the  line  TC  to  the  vertex,  C, 
of  the  triangle. 

2.  From  a  and  b  draw  the 

lines  aD,  bK,  parallel  to  TC,  meeting  the  sides  AC,  BC, 
in  D  and  K. 

3.  Upon  AC,  from  A  towards  C,  measure  off  the  dis* 
tance  AD  as  many  times  as  possible  (in  this  case  four 
times)  ;  and  thus  determine  the  points  E,  F,  G ;  upon 
BC,  in  the  direction  from  B  towards  C,  also  measure  off 
the  distance  BK,  as  many  times  as  is  possible  (here  three 
times),  and  determine  the  points  I,  H. 

4.  From  T  draw  the  lines  TD,  TE,  TF,  TG,  TH, 
TI,  TK  ;  then  ATD,  DTE,  ETF,  FTG,  GTHC,  HTI, 
ITK,  KTB,  are  the  eight  equal  parts  of  the  triangle  ABC. 


GEOMETRY. 


173 


Demon.  Draw  Ca;  then  the  triangle  AaC  is  \  of  the  triangle 
ABC  ;  because,  if  AB  is  taken  for  the  base  of  the  triangle  ABC,  the 
base  Aa  of  the  triangle  AaC  is,  by  construction,  §  of  the  base  of 
the  triangle  ABC  (see  page  171).  Now,  the  triangle  aDC  is  equal 
to  the  triangle  aDT  ;  because  these  two  triangles  are  upon  the 
same  base,  aD,  and  between  the  same  parallels,  aD,  TC  ;  therefore 
(by  adding  to  each  of  them  the  triangle  aAD)  the  two  triangles 
ADT  ancfaAC  are  also  equal ;  that  is,  ADT  is  also  J  of  the  trian- 
gle-ABC. In  the  same  manner  (by  drawing  the  line  6C)  it  may 
be  proved  that  the  triangle  BKT  is  also  J  of  the  triangle  ABC. 
Further,  the  triangles  ATDa  DTE,  ETF,  FTG,  are,  by  construc- 
tion, all  equal  to  one  another,  having  equal  bases  and  heights  (see 
the  demonstration  to  the  last  problem)  ;  and  for  the  same  reason  are 
the  triangles  BTK,  KTI,  ITH  equal  to  one  another ;  therefore  each 
of  the  seven  triangles  ATD,  DTE,  ETF,  FTG,  BTK,  KTI,  ITH,is 
£  of  the  triangle  ABC  ;  consequently  the  quadrilateral  GTHC 
must  be  the  remaining  one  eighth  of  the  triangle  ABC;  and  the 
area  of  the  triangle  ABC  is  divided  into  eight  equal  parts. 


Problem   XXXVI.       To  divide  a  triangle,  from  a 
given  point  within  it,  into  a  given  number  of  equal  parts. 
[This  problem  is  intended  for  elder  pupils.] 
A. 


Solution.  Let  ABC  be  the  given  triangfe,  which  is 
to  be  divided,  say,  into  five  equal  parts  ;  T  the  point 
from  which  the  lines  of  division  are  to  be  drawn. 

1.  Through  the  point  T  and  the  vertex  A  of  the  trian- 
gle, draw  the  line  AT. 
15* 


174  GEOMETRY. 

2.  Take  any  side  of  the  triangle,  say  BC,  and  make, 
when,  as  here,  the  triangle  is  to  be  divided  into  five  equal 
parts,  BE  and  CF  equal  to  £  of  BC,  and  draw  the  lines 
Ee,  'Ff,  parallel  to  the  sides  AB,  AC  ;  these  lines  will 
meet  the  line  AT  in  the  points  e  and/". 

3.  From  T  draw  the  lines  TB,  TC,  to  the  vertices  B 
and  C  of  the  triangle  ABC^  and  from  e  and/,  the  lines 
eI,fG  parallel  to  TB,  TC/ 

4.  Join  TI,  TG  ;  then  each  of  the  triangles  ATI, 
ATG,  is  i  of  the  given  triangle  ABC. 

5.  In  order  to  determine  the  other  points  of  division, 
it  is  only  necessary  to  cut  off  from  the  sides  AB,  AC,  as 
many  distances,  equal  to  AI,  AG,  respectively,  as  is  possible 
(see  the  solution  of  the  last  problem),  and  in  the  case  where 
this  can  no  longer  be  effected,  or  in  which,  as  in  the  figure, 
this  is  impossible,  proceed  in  the  following  manner  : 

a.  Extend  the  two  sides  AB,  AC,  and  then  make  IM 
equal  to  AI,  and  GN  equal  to  AG. 

b.  From  M  and  N  draw  the  lines,  MH,  NP,  parallel  to 
BT  and  CT,  and  determine  thereby  the  points  H  and  P. 

6.  Draw  TH,  TP ;  each  of  the  quadrilaterals  IBHT, 
GCPT,  is  £  of  the  triangle  ABC  ;  consequently  the  trian- 
gle HTP  is  the  remaining  fifth  of  it.  (If  HTB  were  not 
the  last  part,  then  it  would  merely  be  necessary  to  divide 
this  triangle  by  the  rule  given  in  problem  XXXIV,  into 
as  many  equal  parts  as  necessary.) 


DfiMON.  Draw  the  auxiliary  lines  AE,  Be  ;  then  the  triangle 
ABE  is  one  fifth  of  the  triangle  ABC ;  because  BE  is  one  fifth  of 
the  basis  BC  (problem  XXXIV) ;  further,  the  triangle  ABE  is 
equal  to  the  triangle  ABe;  because  these  two  triangles  are  upon 
the  same  basis,  AB,  and,  by  construction  2,  between  the  same  paral- 
teh,  AB,  Ee;  and  the  last  triangle,  ABe,  is  also  equal  to  the  triangle 
ATI ;  because  the  triangle  ABe  consists  of  the  two  triangles  Ale 
and  IeB,  which  are  equal  to  the  two  triangles  Ale  and  ITe  (the 


GEOMETRY.  175 

two  triangles  ITe  and  IeB  being  upon  the  same  base,  Je,  and,  by- 
construction  3,  between  the  same  parallels,  Ie,  BT) ;  therefore  the 
triangle  AIT  is  also  one  fifth  of  tbe  triangle  ABC  ;  and  in  the  same 
manner  it  can  be  proved  that  ATG  is  one  fifth  of  the  triangle 
ABC. 

Further,  the  triangle  GNT  is  equal  to  the  triangle  A  GT  (the 
basis  GN  being  made  equal  to  the  basis  AG,  and  the  vertical  point 
T  being  common  to  both  triangles)  ;  and  the  triangle  GNT  is  equal 
to  the  quadrilateral  CGTP ;  because  the  triangle  CTN  is  equal  to 
the  triangle  CTP  (these  two  triangles  being,  by  construction,  upon 
the  same  base,  TC,  and  between  the  same  parallels,  TC,  PN)  ;  there- 
fore the  area  of  the  quadrilateral  CGTP  is  also  one  fifth  of  the  tri- 
angle ABC ;  and  in  the  same  manner  it  may  be  proved  that  the 
area  of  the  quadrilateral  IBHT  is  one  fifth  of  the  triangle  ABC ; 
and  as  the  two  triangles  AGT,  AIT,  together  with  the  two  quad- 
rilaterals CGTP,  IBHT,  make  four  fifths  of  the  triangle  ABC,  the 
triangle  HPT  must  be  the,  remaining  one  fifth  of  it. 

Problem  XXXVII.  To  divide  a  given  triangle  into 
a  given  number  of  equal  parts,  and  in  such  a  way,  that 
the  lines  of  division  shall  be  parallel  to  a  given  side  of  the 
triangle. 

d 


Solution.  Let  ABC  be  the  given  triangle ;  let  the 
number  of  the  parts,  into  which  it  is  required  to  be  di- 
vided, be  five,  and  BC  the  side  to  which  the  lines  of 
division  are  to  be  parallel. 

1.  Upon  one  of  the  other  two  sides,  say  AC,  describe 
a  semicircle,  and  divide  the  side  AC  into  as  many  equal 


176  GEOMETRY. 

parts  as  the  triangle  is  to  be  divided  into;  consequently, 
in  the  present  case,  into  five  ;  the  points  of  division  are 
D,  E,  F,  G. 

2.  From  these  points  of  division  draw  the  perpendicu- 
lars Dd,  Ee,  Ff,  Gg,  meeting  the  semicircle  in  the  points 

3.  From  A  draw  Ad,  Ae,  Af,  Ag ;  then  make  Am 
equal  to  Ad,  An  equal  to  Ae,  and  so  on,  and  by  these 
means  determine  the  points  m,  n,  o,  p. 

4.  From  these  points  draw  the  lines  mM,  wN,  oO,  pV, 
parallel  to  the  side  BC ;  then  AMm,  MroNn,  NwO<?, 
OoVp,  P/)BC  are  the  five  equal  parts  of  the  triangle  ABC, 
which  were  sought. 

Demon.  Imagine  the  line  dC  drawn  ;  the  triangle  AdC,  in- 
scribed in  the  semicircle,  is  right-angled  in  d;  consequently  we 
have  the  proportion 

AD  :  Ad  =  Ad  :  AC  ; 
and  as,  in  every  geometrical  proportion,  the  product  of  the  mean 
terms  is  equal  to  that  of  the  extremes, 

AdxAd  =  ADxAC; 
consequently,  also, 

Am  X  Am  =  AD  X  AC 
(because  Am  is  made  equal  to  Ad). 

Further,  the  triangles  AMm,  ABC,  are  similar,  because  the  line 
Mm  is  drawn  parallel  to  the  side  BC  in  the  triangle  ABC  (Query 
16,  Sect.  II.)  ;  and  as  the  areas-  of  similar  triangles  are  to  each 
other  as  the  areas  of  the  squares  upon  the  corresponding  sides 
(Query  8,  page  97),  we  have  the  proportion 

triangle  ABC  :  triangle  AMm  =  AC  X  AC  :  Am  X  Am  ; 
therefore,  also, 

triangle  ABC  :  triangle  AMm  =  AC  X  AC  :  AC  X  AD 
(because  Am  X  Am  is  equal  to  AC  X  AD). 

The  last  proportion  expresses,  that  the  area  of  the  triangle  ABC  is 
as  many  times  greater  than  the  area  of  the  triangle  AMm,  as  AC 
times  the  side  AC  itself  is  greater  than  AC  times  the  side  AD;  or, 
which  is  the  same,  as  AC  is  greater  than  AD  (Prin.  7th  of  Geom. 


GEOMETRY.  177 

l>rop.  page  63).  But  the  side  AD  is,  by  construction,  one  fifth  of 
AC  ;  therefore  the  area  of  the  triangle  AMm  is  also  one  fifth  of  the 
area  of  the  triangle  ABC.  In  like  manner  it  may  be  proved,  that 
the  triangle  ANra  is  two  fifths  of  the  triangle  ABC  ;  the  triangle 
AOo  three  fifths,  and  the  triangle  APp  four  fifths  of  it,  from  wbich 
the  rest  follows  of  course. 

Remark.  If  the  triangle  ABC  is  not  to  be  divided  into  -<ual 
parts,  but  according  to  a  given  proportion,  it  will  merely  be  d  pes- 
sary, as  may  be  readily  seen  from  the  above,  to  divide  the  lint  \C 
according  to  this  proportion,  and  then  proceed  as  has  been  ah>*uy 
shown. 

Problem  XXXVIII.  To  divide  a  parallelogram  *nto 
a  given  number  of  equal  parts,  and  in  such  a  way,  t%at 
the  lines  of  division  may  be  parallel  to  two  opposite  sides 
of  the  parallelogi^am. 

Solution.      Let   ABCD  B  C   f   </  h    i    C 

be  the  given  parallelogram  ;  / 


let  the  number  of  parts  be  /   / 

six  ;  ami  let  AS,  CD,  be  the  I  I  I  I  I  I  I 
sides,  to  which  the  lines  of  A  EE  GUI  B 
division  shall  be  parallel. 

Divide  one  of  the  two  other  sides,  say  AD,  into  six 
equal  parts,  in  E,  F,  G,  H,  I,  and  from  these  points  draw 
the  lines  Ee,  Yf,  Gg,  H//,  I*,  parallel  to  the  sides  AB, 
CD ;  then  the  division  is  done. 

Remark.  If  it  is  required  to  divide  the  parallelogram  according 
to  a  given  proportion,  it  will  merely  be  necessary,  instead  of  divid- 
ing the  line  AD  into  equal  parts,  to  divide  it  according  to  the  given 
proportion,  and  then  proceed  as  before. 

Problem  XXXIX.  To  divide  a  parallelogram,  ac- 
cording to  a  given  proportion,  by  a  line  which  shall  be 
parallel  to  a  line  given  in  position. 


178 


GEOMETRY. 


Aa  *     E 


D 


Solution.  Let  ABCD  be  the  parallelogram  to  be 
divided. 

1.  Divide  one  of  its  sides,  say  AD,  according  to  the 
given  proportion  ;  let  the  point  of  division  be  in  z. 

2.  Make  zE  equal  to  the  distance  Az,  and  draw  BE. 
Now,  if  the  line  BE  has  the  required  position,  the  tri- 
angle ABE  and  the  quadrilateral  BCDE  are  the  parts 
sought. 

3.  But  if  the  line  of  division  is  required  to  be  parallel 
to  the  line  xy,  bisect  the  line  BE  in  t,  and  through  this 
point  draw  the  line  GH  parallel  to  xy ;  then  the  two 
quadrilaterals  ABHG,  HCDG,  will  be  the  required  parts. 

Demox.  Draw  EK  parallel  to  AB.  Then  the  two  parallel- 
ograms ABEK,  ABCD,  having  the  same  height,  their  areas  are 
in  proportion  to  their  hases,  AE,  AD  (see  page  88,  7th)  ;  that  is, 
we  have 

parallel.  ABEK  :  parallel.  ABCD  =  AE  :  AD ; 
therefore 

-   £  of  parallel.  ABEK  :  parallel.  ABCD  =  £  of  AE  :  AD ; 
and  because  the  triangle  AEB  is  equal  to  half  the  parallelogram 
ABEK,  and  half  of  AE  is,  by  construction,  equal  to  Az,  we  have 
triangle  AEB  :  parallel.  ABCD  =  Az  :  AD. 

This  last  proportion  expresses  that  the  area  of  the  parallelogram 
ABCD  is  as  many  times  greater  than  the  area  of  the  triangle 
ABE,  as  the  line  AD  is  greater  than  Az  ;  consequently  if  BE  has 
the  required  position,  the  triangle  ABE  is  one  of  the  required 
parts,  and  therefore  the  trapezoid  BEDC  the  other. 

Further,  the  line  BE  is  (by  construction  3)  bisected ;  the  an- 


GEOMETRY, 


179 


gles  u  and  n  are  opposite  angles  at  the  vertex,  and  w  and  s 
are  alternate  angles  (page  31,  2d) ;  therefore  the  triangle  BtH, 
having  the  side  Bt,  and  the  two  adjacent  angles,  to  and  u,  equal 
to  the  side  tE,  and  the  two  adjacent  angles,  n  and  s,  in  the 
triangle  GtE,  these  two  triangles  are  equal  to  one  another  ;  conse- 
quently the  area  of  the  trapezoid  ABGH  (composed  of  the  quadri- 
lateral ABGt,  and  the  triangle  B/H)  is  equal  to  the  area  of  the 
triangle  ABE  (composed  of  the  same  quadrilateral  ABGt  and 
the  equal  triangle  GtE),  which  proves  the  correctness  of  con- 
struction 3. 

Problem  XL.  To  divide  a  trapezoid  into  a  given 
number  of  equal  parts,  so  that  the  lines  of  division  may 
be  parallel  to  the  parallel  sides  of  that  trapezoid. 

[This  problem  may  be  omitted  by  the  younger  pupils.] 


Solution.  Let  ABCD  be  the  given  trapezoid  which 
is  to  be  divided  into  three  equal  parts. 

1.  Upon  AB,  the  greater  of  the  two  parallel  sides,  de- 
scribe a  semicircle ;  draw  DE  parallel  to  CB ;  and  from 
B,  with  the  radius  BE,  describe  the  arc  of  a  circle,  EF, 
cutting  the  semicircle  in  F. 

2.  From  F  draw  FG  perpendicular  to  AB,  and  divide 


180  GEOMETRY. 

the  part  AG  of  the  line  AB,  into  three  equal  parts 
in  K  and  I ;  from  these  points  draw  the  perpendiculars 
Kk,  li. 

3.  Upon  AB,  from  B  towards  A,  take  the  distances 
Bm,  Bn,  equal  to  Bk>  Bi ;  from  the  points  m  and  n,  draw 
the  lines  mO,  wM,  parallel  to  BC ;  and  from  the  points 
O,  M,  in  which  these  parallels  meet  the  side  AD,  the 
lines  MN,  OP,  parallel  to  AB :  then  ABNM,  MNPO, 
OPCD,  are  the  three  required  parts  of  the  trapezoid 
ABCD 

Demon.    Extend  the  lines  AD,  BC,  until  they  meet  in  Z.    Then 
the  triangles  DCZ,  OFZ,  MNZ,  ABZ,  are  all  similar  to  each  other 
(page  70)  ;  further,  we  have  (by  construction  3) 
DC  equal  to  BE  and  to  BF, 
OP      «      «  Bra   "    "    Bk, 
MN    «      "   Bn    "    «   Bi. 
The  areas  of  the  two  similar  triangles  OPZ,  CDZ,  are  in  the  ratio 
of  the  squares  upon  the  corresponding  sides;  that  is,  we  have  the 
proportion 

triangle  OPZ  :  triangle  DCZ  =  OP  X  OP  :  CD  X  CD ; 
and  since  OP  is  equal  to  Bk,  and  CD  to  BF,  also 

triangle  OPZ  :  triangle  DCZ  =  Bk  X  B/e  :  BF  X  BF. 
Imagine  AF  and  FB  joined  ;  the  triangle  AFB  would  be  right- 
angled  in  F,  and  we  should  have  the  proportion 

BG  :  BF  =  BF  :  AB ; 
and  for  the  same  reason  we  have 

BK  :  Bk  =  Bk:  AB. 
Taking  the  product  of  the  mean  and  extreme  terms  of  the  two 
lastpropoitions,  we  have 

BG  X  AB  equal  to  BF  X  BF,  and 
BK  x  AB      «     «    BA  X  Bk 
Let  us  now  take  our  first  proportion, 

triangle  OPZ  :  triangle  DCZ  =  Bk  X  Bk  :  BF  X  BF  ; 
and  let  us  write  BG  X  AB,  instead  of  BF  X  BF   (its  equal),  and 
BK  X  AB,  instead  ofBk  X  B/c,  and  we  shall  have 

triangle  OPZ  :  triangle  DCZ  =  AB  X  BK  :  AB  X  BG, 
whence 

triangle  OPZ  :  triangle  DCZ  =  BK  :  BG; 


GEOMETRY.  181 

consequently,  also, 

triangle  OPZ  — triangle  DCZ  :  triangle  DCZ  =  BK  —  BG  :  UB; 

(Principle  6th  of  Geom.  Prop,  page  62) ;  which  is  read  thus : 

triangle  OPZ,  less  the  triangle  DCZ,  is  to  the  triangle  DCZ, 

as  the  line  BK,  less  the  line  BG,  is  to  the  line  BG; 

that  is,  trapezoid  DOPC  :  triangle  DCZ=  GK  :  BG; 

and  as  GK  is  (by  construction  2)  equal  to  J  of  AG, 

trapezoid  DOPC  :  triangle  DCZ  =  £  AG  :  BG. 
In  like  manner  it  may  be  proved  that 

trapezoid  DMNC  :  triangle  DCZ  =  §  AG  :  BG  and 
trapezoid  DABC  :  triangle  DCZ  =  AG  :  BG. 
These  proportions  express  that  the  three  trapezoids  DOPC, 
DMNC,  DABC,  are  to  each  other  in  the  same  proportion  as  one 
third  is  to  two  thirds  to  three  thirds  ;  or,  which  is  the  same,  as  one 
is  to  two,  to  three ;  whence  the  rest  of  the  demonstration  follows 
of  course. 

Remark.  If  it  is  required  to  divide  the  trapezoid  ABCD  not 
into  equal  parts,  but  according  to  a  given  proportion,  it  will  only 
be  necessary  to  divide  the  line  AG  in  this  proportion,  and  then 
proceed  as  before. 

Problem  XLI.  To  divide  a  given  figure  into  tie* 
parts  according  to'  a  given  proportion,  and  in  such  a  way, 
that  one  of  the  parts  may  be  similar  to  the  zohole  figure. 

D 


Solution.     Let  ABCDE  be  the  given  figure. 

1.  Divide  one  side  of  the  figure,  say  AB,  according  to 
the  given  proportion ;  let  the  point  of  division  be  Z. 

2.  Upon  AB,  as  a  diameter,  describe  a  semicircle,  and 


16 


183  GEOMETRY. 

from  Z  draw  the  perpendicular  ZM,  meeting  the  semi- 
circle in  M. 

3  Make  A6  ±=  AM,  and  upon  A&  describe  a  figure, 
Abcde,  which  is  similar  to  the  given  one,  ABCDE  (see 
Problem  XXXIII) ;  the  line  bcde  divides  the  figure  in 
the  manner  required. 

Demon.  The  areas  of  the  two  similar  figures  Abcde,  ABCDE, 
are  to  each  other,  as  the  squares  upon  their  corresponding  sides 
(page  98)  ;  therefore  we  have  the  proportion 

ABCDE  :  Abcde  =  AB  X  AB  :  A6  X  A&. 
Draw  AM  and  BM  ;  then  AM  is  a  mean  proportional  between 
AZ  and  AB  ;  that  is,  we  have 

AZ  :  AM  =  AM  :  AB ; 
and  as  A6  is,  by  construction,  equal  to  AM, 
AZ:  Aft  — A&:  AB; 
consequently  the  product  A6  X  A6  is  equal  to  AZ  X  AB. 

Writing  AZ  X  AB,  instead  of  Aft  X  Aft  (its  equal),  in  the  first 
proportion,  we  have 

ABCDE  :  Abcde  =  AB  X  AB  :  AB  X  AZ. 
Hence  ABCDE  :  Abcde  =  AB  :  AZ ;  and  therefore 

ABCDE  —  Abcde  :  Abcde  =  AB  —  AZ  :  AZ ; 
which  is  read  thus  : 

ABCDE,  less  Abcde,  is  to  Abcde  as  AB,  less  AZ,  is  to  AZ ; 
that  is, 

BCDEedcb  is  to  Abcde  as  ZB  is  to  AZ ; 
consequently  the  figure  ABCDE  is  divided  according  to  the  given 
proportion  in  which  the  line  AB  is  divided. 


PART  IV. 

Construction  of  triangfe$. 

Problem  XLII.     The  three  sides  of  a  triangle  being 
given,  to  construct  the  triangle. 


GEOMETRY. 


183 


AAB 


Solution.  Let  AB,  AC, 
BC,  be  the  three  given  sides 
of  the  triangle. 

1.  Take  any  side,  say 
AB.  and  from  A  as  a  centre, 
with  the  radius  AC,  describe 
an  arc  of  a  circle. 

2.  From  B,  as  a  centre,  with  the  radius  BC,  describe 
another  arc,  cutting  the  first. 

4.  From  the  point  of  intersection  C,  draw  the  straight 
lines  CA,  CB  ;  the  triangle  ABC  is  the  one  required. 

The  demonstration  follows  immediately  from  Query  4th,  Sect.  II. 


B 


Problem  XLIII.     Two  sides,  and  the  angle  included 
by  them,  being  given,  to  construct  the  triangle. 


A  A 


Solution.  Let  AB,  AC,  be  the  two  given  sides,  and 
x  the  angle  included  by  them. 

1.  Construct  an  angle  equal  to  the  angle  x  (Problem 
VI) ;  make  one  of  the  legs  equal  to  the  side  AB,  and  the 
other  to  the  side  AC. 

2.  Join  BC  ;  the  triangle  ABC  is  the  one  required. 

The  demonstration  follows  from  Query  1,  Sect  II. 


184 


GEOMETRY. 


Problem    XLI V.     One  side   and  the   two   atfyaesnt 
angles  being  given,  to  construct  the  triangle 


Solution.  Let  AB  be  the  given  side,  and  x  and  y 
the  two  adjacent  angles. 

1.  At  the  two  extremities  of  the  line  AB,  construct 
the  angles  x  and  y,  and  extend  their  legs,  AC,  BC,  until 
they  meet  in  the  point  C ;  the  triangle  ABC  is  the  one 
required. 

The  demonstration  follows  from  Query  2,  Sect.  II. 


Problem  XLV.  Two  sides,  and  the  angle  opposite  to 
the  greater  of  them,  being  given,  to  construct  the  triangle. 

Solution.  Let  AC,  BC  (see  the  figure  to  Problem 
XLIII)  be  two  given  sides,  and  x  the  angle,  which  is 
opposite  to  the  greater  of  them  (the  side  BC). 

1.  Upon  an  indefinite  straight  line  construct  an  angle 
equal  to  the  angle  x. 

2.  Make  the  leg  AC  of  this  angle  equal  to  the  smaller 
side  AC,  and  from  C  as  a  centre,  with  the  radius  GB 
equal  to  the  greater  side,  describe  an  arc  of  a  circle, 
cutting  the  line  AB  m  the  point  B. 

3.  Join  BC  ;  the  triangle  ABC  is  the  one  required. 

The  demonstration  follows  from  Query  10th,  Sect.  11. 


GEOMETRY. 


185 


Problem  XLVI.  The  basis  of  a  triangle,  one  of  the 
adjacent  angles,  and  the  height  being  given,  to  construct 
the  triangle. 

c 

M  EC 


BB 


Solution.     Let  AB  be  the  given  basis,  x  one  of  the 
adjacent  angles,  and  cd  the  height. 

1.  In  any  point  of  the  line  AB,  draw  a  perpendicular, 
CD,  equal  to  cd ;  and  through  C  a  line  parallel  to  AB. 

2.  In  A  make  an  angle  equal  to  the  given  angle  x, 
and  extend  the  leg  AE  until  it  meets  the  line  MC. 

3.  Join  EB  ;  the  triangle  AEB  is  the  one  required. 

The  demonstration  is  sufficiently  evident  from  the  construction. 


Problem  XLVII.  The  basis,  the  angle  opposite  to  it9 
and  the  height  of  a  triangle  being  given,  to  construct  the 
triangle. 


As^ 


m 


Solution.     Let  AB  be  the  given  base,  x  the  angle 
opposite  to  it,  and  ran  the  height  of  the  triangle. 

1 .  Upon  the  base  AB  describe  a  segment  of  a  circle 
containing  a  given  angle  x  (see  Problem  XVII). 

2.  In  A  draw  a  perpendicular,  AD,  equal  to  the  given 
height  mn,  and  through  D  draw  DE  parallel  to  AB. 

16* 


186 


GEOMETRY. 


3.  From  C  and  E,  where  this  parallel  cuts  the  segment, 
draw  the  straight  lines  CA,  CB,  EA,  BE  ;  either  of  the 
two  triangles  ACB,  AEB,  will  be  the  one  required. 

The  demonstration  follows  from  the  construction. 

Problem  XLVIII.  The  basis  of  a  triangle,  the  angle 
opposite  to  it,  and  the  ratio  of  the  two  other  sides  being 
given,  to  construct  the  triangle. 


Solution.  Let  AB  be  the  given  basis,  x  the  angle 
opposite  to  it ;  and  let  the  two  remaining  sides  bear  to 
each  other  the  same  ratio  which  exists  between  the  two 
lines  mn  and  rq. 

1.  Upon  AB  describe  a  segment  of  a  circle  capable 
of  the  given  angle  x  (see  Problem  XVII). 

2.  In  B  make  an  angle,  ABE,  equal  to  the  angle  x  ; 
make  BE  equal  to  the  line  rq,  BD  equal  to  mn,  and  join 
DE. 

3.  From  A  draw  the  line  AC  parallel  to  DE,  and  from 
the  point  C,  where  it  meets  the  segment,  draw  the  line 
CB ;  the  triangle  ABC  is  the  one  required. 

Demon.  The  triangle  ABC  is  similar  to  the  triangle  DBE ; 
because  the  two  angles  CAB  and  ACB,  in  the  one,  are  equal  to  the 
two  angles  BDE,  DBE,  in  the  other,  each  to  each*  (page  73, 1st) ; 
therefore  we  have  the  proportion 

*  CAB  and  EDB  being  alternate  angles,  and  each  of  the  angles, 
ACB,  DBE,  being  made  equal  to  the  given  angle  x. 


\M. 


ft 


GEOMETRY. 


187 


AC  :  BC  =  BE  :  BD, 

which  expresses  that  the  two  sides,  AC,  BC,  of  the  triangle  are  in 
the  same  ratio  as  the  sides  BE,  BD,  of  the  triangle  DEB  ;  conse- 
quently they  are  also  as  the  lines  rq,  mn ;  because  BE  and  BD 
are,  by  construction,  equal  to  mn,  rq.  The  rest  of  the  demonstra- 
tion is  evident  from  the  construction. 


Problem  XLIX.  The  basis  of  a  triangle,  the  angle 
opposite  to  it,  and  the  square,  which,  in  area,  is  equal  to 
the  rectangle  of  the  two  remaining  sides,  being  given,  to 
construct  the  triangle.* 

[Let  the  younger  pupils  omit  this  problem.] 


Solution.  Let  AB  be  the  given  base,  x  the  angle 
opposite  to  it,  and  ad  the  side  of  the  square,  equal  to  the 
rectangle  of  the  two  remaining  sides. 

1.  Upon  AB  construct  the  segment,  AKHB,  of  a  cir- 
cle, capable  of  the  given  angle  x. 

2.  Extend  AB  towards  D,  and  in  A  draw  the  perpen- 
dicular AF. 

3.  Make  AC  equal  to  the  radius  AO,  AD  to  the  side 
ad  of  the  given  square,  and  AE  to  half  of  AD  ;  join  EC, 
and  from  D  draw  DF  parallel  to  EC. 

4.  Through  the  point  F,  where  this  parallel  meets  the 
perpendicular,  draw  FH  parallel  to  AB ;  and  from  the 
points  K  and  H,  where  this  meets  the  segment,  the  lines 


*  By  the  rectangle  of  the  two  remaining  sides  is  meant  a  rectan- 
gle, whose  base  is  one  of  these  sides,  and  whose  height  is  the  other. 


188  GEOMETRY. 

AK,  KB,  AH,  HB;  then  either  of  the  two  triangles  AKB, 
AHB,  is  the  one  required. 

Demojv.  Draw  the  diameter  AL,  and  from  either  of  the  points 
K,  H,  say  H,  let  fall  the  perpendicular  HM  upon  AB.  The  trian- 
gle ALH  is  similar  to  the  triangle  MBH  ;  for  the  triangle  ALH 
being  inscribed  in  a  semicircle,  each  of  these  triangles  is  right- 
angled,  and  the  two  angles  ALH,  ABH,  are  equal ;  because  both 
of  them  measure  half  as  many  degrees  as  the  arc  AKH  (page  111, 
1st);  therefore  the  remaining  angles,  HAL  and  MHB,  are  also 
equal  (page  73,  1st)  ;  and  the  corresponding  sides  of  the  two  trian- 
gles ALH,  MBH,  are  in  the  geometrical  proportion 

AH  :  AL  =  HM  :  HB  ; 
consequently  we  have 

ALX  HM=AHxHB. 

This  proportion  expresses,  that  the  area  of  the  rectangle,  which 
has  for  its  base  the  diameter  AL,  and  its  height  equal  to  the  height 
HM  of  the  right-angled  triangle  AHB,  is  equal  to  the  area  of  the 
rectangle,  which  has  the  side  AH  for  its  base,  aid  the  side  HB 
for  its  height.*  Further,  it  is  easy  to  perceive  that,  from  the  similar 
triangles  ACE,  ADF,  we  have  the  proportion 
AD  :  AF  =  AC  :  AE ; 
consequently,  also, 

AD  :  AF  =  2AC  :  2AE , 
therefore, 

2AC  X  AF  =  2AE  X  AD  ;  or 
diam.  AL  X  AF  =  adX  ad, 
(because  AC  is  equal  to  the  radius  Ao  of  the  circle,  and  AE  is  half 
of  AD,  and  AD  is  equal  to  ad). 

From  this  proportion  it  follows,  that  the  area  of  the  square  upon 
ad,  is  equal  to  that  of  the  rectangle  of  AL  by  AF,  or  MH  its  equal 
(see  the  figure)  ;  and  as  the  rectangle  AH  by  HB  is  equal  to  that 
of  AL  by  HM,  as  we  have  proved  above,  it  must  also  be  equal  to 
the  square  upon  ad.  The  same  may  be  proved  of  the  rectangle  of 
the  two  sides  AK,  KB,  of  the  triangle  AKB. 

The  rest  of  the  demonstration  is  sufficiently  evident  from  the 
construction. 

*  For  the  area  of  a  rectangle  is  found  by  multiplying  the  base  by 
the  height. 


GEOMETRY.  igg 

APPENDIX. 

Containing  Exercises  for  the  Slate. 

1.  The  side  of  a  square  being  12  feet,  what  is  its  area! 

2.  What,  if  the  side  is  12  rods,  miles,  &c? 

3.  What  is  the  side  of  a  square,  whose  area  is  one 
square  foot? 

4.  What,  that  of  a  square,  whose  area  is  one  square 
yard,  rod,  mile,  &c.  ? 

5.  What,  that  of  a  square  of  4,  9,  16,  25,  36,  49,  64, 
81,  lOOsqare  feet? 

6.  What  is  the  area  of  a  rectangle,  whose  base  is  50 
feet  3  inches,  and  whose  height  10  feet  4  inches  ? 

7.  What,  that  of  a  rectangle,  whose  base  is  40  feet  3 
inches,  and  whose  height  is  12£  feet  ? 

8.  If  the  area  of  a  rectangle  is  240  square  feet  19 
square  inches,  and  its  basis  measures  30  feet,  what  is  its 
height  ? 

9.  What  is  the  basis  of  a  rectangle,  whose  height  is 
10  feet,  and  whose  area  is  40  square  feet  ? 

10.  What  is  the  area  of  a  rectangle,  whose  basis  is  4 
feet,  and  whose  height  is  3  inches  ? 

11.  What  is  the  area  of  a  parallelogram  of  10  feet 
basis,  and  3  feet  4  inches  high  ? 

12.  The  height  of  a  parallelogram  is  5  feet,  and  the 
area  40  square  feet :  what  is  its  basis  ? 

13.  The  sum  of  the  two  parallel  sides  of  a  trapezoid  is 
12  feet,  and  their  distance  3  feet  4  inches :  what  is  the 
area  of  the  trapezoid  ? 

14.  The  area  of  a  trapezoid  is  24  square  feet,  and  its 
height  is  4  inches,  3  seconds :  what  is  the  sum  of  its  bases  1 


190  GEOMETRY. 

15.  What  is  the  difference  between  a  triangle  whose 
basis  is  10  feet  3  inches,  and  height  9  feet,  and  a  trian* 
gle  of  3  feet  basis,  and  11  inches  height? 

16.  What  is  the  difference  between  a  trapezoid,  the 
sum  of  the  two  parallel  sides  of  which  is  14  feet  3  inches, 
and  height  9  inches,  and  a  square  upon  9  inches'? 

17.  What  is  the  sum  of  the  areas  of  a  triangle  of  3  feet 
basis,  and  9  inches  height;  a  square  upon  14  feet  3 
inches,  and  a  rectangle  whose  basis  is  3  feet  2  inches, 
and  height  1  foot  4  inches  ? 

18.  What  is  the  area  of  a  circle,  whose  radius  is  9 
inches  1 

19.  What  that  of  a  circle,  whose  radius  is  10  feet? 

20.  What  that  of  a  circle,  whose  radius  is  9  feet  6 
inches  ? 

21.  The  area  of  a  circle  is  240  square  feet :  what  is  its 
radius  or  diameter?* 

22.  The  radius  of  a  circle  is  5  feet  8  inches  :  what  is 
its  circumfereuce  ? 

23.  What  is  the  length  of  an  arc  of  14  degrees  29 
minutes  24  seconds,  in  a  circle  whose  radius  is  14  inches  ? 

24.  What  that  of  an  arc  of  6  degrees  9  seconds,  in  a 
circle  whose  radius  is  1  foot  ? 

25.  What  that  of  an  arc  of  9  seconds,  in  a  circle  whose 
radius  is  1  mile  ? 

20.  What  is  the  area  of  a  sector  of  15  degrees,  in  a 
circle  whose  radius  is  3  feet  ? 

27.  What  that  of  a  sector  of  19  degrees  45  minutes,  in 
a  circle  whose  radius  is  1  foot  3  inches  ? 

The  teacher  may  now  vary  and  multiply  these  questions. 

*  Divide  the  area  by  n  (see  the  note  to  page  132),  and  extract 
the  square  root  of  the  quotient,  the  answer  is  the  radius  of  the 
circle. 

END. 


* 


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